At each step of a limiting infinite process, put 10 balls in an urn and remove one at random. How many balls are left? The question (slightly modified) goes as follows and if you have never encountered it before you can check it in example 6a, chapter 2, of Sheldon Ross' A First Course in Probability:

Suppose that we possess an infinitely large urn and an infinite
  collection of balls labeled ball number 1, number 2, number 3, and so
  on. Consider an experiment performed as follows: At 1 minute to 12
  P.M., balls numbered 1 through 10 are placed in the urn and one ball
  removed at random. (Assume that the withdrawal takes no time.) At 1/2
  minute to 12 P.M., balls numbered 11 through 20 are placed in the  urn
  and another ball removed at random. At 1/4 minute to 12P.M., balls
  numbered 21 through 30 are placed in the urn and another ball removed
  at random... and so on. The question of interest is, How many balls
  are in the urn at 12 P.M.?

This question, as it's posed, forces basically everyone to get it wrong --- usually the intuition is to say there will be infinitely many balls at 12 P.M. The  answer provided by Ross, however, is that with probability one the urn will be empty at 12 P.M.
When teaching probability theory this problem is one of those for which is very hard to give a good intuitive explanation.  
On the one hand, you could try to explain it like this: "think of the probability of any ball i being on the urn at 12 P.M. During the infinite random draws, it will eventually be removed. Since this holds for all balls, none of them can be there at the end". 
However, students will correctly argue with you: "but I'm putting 10 balls and removing 1 ball at each time. It's impossible there will be zero balls at the end".
What's the best explanation we can give to them to solve these conflicting intuitions?
I'm also open to the argument the question is ill-posed and that if we formulate it better the "paradox" disappears or to the argument that the paradox is "purely mathematical"  (but please try to be precise about it).
 A: The problem is either ill-formed or not in first-order logic.
Root cause: execution of the "last" step will write an infinite number of digits on a ball, causing that step to take itself an infinite time to execute.
The ability to execute an infinite process with an infinite step implies the ability to solve all first-order logic problems (Gödel is therefore false) by execution of the following sequence H (for theorem X):
Z = asymptotic_coroutine(
  FOR N = 1...∞
    FOR P = 1...N
      Convert number P to string S by characters.
      IF S is a proof for theorem X
      THEN
        OUTPUT "yes" and HALT
) + asymptotic_coroutine(
  FOR N = 1...∞
    FOR P = 1...N
      Convert number P to string S by characters.
      IF S is a proof for theorem ¬X
      THEN
        OUTPUT "no" and HALT
)
IF Z = "" 
THEN Z = "independent"
IF Z = "yesno" ∨ Z = "noyes"
THEN Z = "paradox"
OUTPUT Z

where the infinite step is unspooling the output
The program inside the asymptotic_coroutine is merely an exhaustive search for a theorem that proves (or disproves) X. Converting P to S results in "aa", "ab", "ac", ... "a∨", ... where every symbol that can appear in a theorem is generated. This results in generating all theorems of length logcharacters N in turn. Since N grows without limit in the outer loop this will eventually generate all theorems.
The side that is false will never terminate but we don't have to care about that because we are allowed to execute infinite steps. In fact we depend on being able to do this to detect independence as both sides will never finish. Except for one thing. We allowed an infinite number of steps to execute in a finite time by asymptotic increase of execution speed. This is the surprising part. The asymptotic_coroutine that never finishes and never generates output has "finished"* after the asymptotic time and still has never generated any output.
*If we placed an OUTPUT after the FOR N = 1...∞ it would not be reached but we are not going to do that.
The strong form of Gödel's Incompleteness Theorem may be stated "For every first-order logic system F there is a statement GF that is true in F but cannot be proven to be true in F." But proof method H cannot fail to prove all must-be-true statements in F(H).
Dilemma: ¬Gödel ∨ ¬(infinite steps are allowed)
Therefore:
Dilemma: ¬Gödel ∨ ¬(315502 is well formed in first order logic)
A: 
What's the best explanation we can give to them to solve these
  conflicting intuitions?

Here's the best answer, and it has very little to do with probabilities. All balls have numbers, let's call them birth numbers. The birth numbers start from B1, B2, B3... and go to infinity, because we really never stop. We get closer to 12:00AM but keep adding and removing balls, that's why there is not a final number of a ball. This is a very important consideration, btw.
We put the balls into a box in 10 ball batches, such as batch #7: B71, B72,...,B80. Let's forget about these for a minute, and focus on the balls that are removed from the box. They come at a random order. I'll explain why randomness is important later, but for now all it means is that any ball with a brith number from B1 to B10k that's still in the box at step K can be drawn out.  We're going to index the balls that we remove by the order in which they were removed, let's call them death numbers: D1, D2, D3 ... DK. 
By 12:00AM we put infinite number of balls into a box, and surely we never ran out of balls to remove from it. Why? Because we first put 10 balls, THEN ONLY remove one. So, there's always a ball to remove. This means that we also removed infinite number of balls by 12:00AM.
This also means that each removed ball was indexed from 1 to infinity, i.e. we could pair each removed ball to a ball that was put in the box: B1 to D1, B2 to D2, etc. This means that we removed as many balls as we put in, because each birth number was paired with each death number.
Now that was the solution. Why does it defeat our intuition? It's elementary, Dr Watson. The reason is because we surely know that for all K this holds:
$$K<10K$$
That's why after K steps, we should not be able to remove all ball from the box, because we put 10K balls and removed only K of them. Right? 
There is a little problem. The matter is that when $K=\infty$, this is no longer true:
$$10\times\infty\nless\infty$$
That's why the intuition breaks down.
Now, if the balls were not removed at random. Two thing may happen as in @amoeba's canonical answer. First, let's say we were putting 10 balls then immediately removing the last one. It's as if we were putting only nine balls in. This will match our intuition, and at 12:00AM there will be infinite number of balls. How come? Because we were not removing balls randomly, we were following the algorithm where the birth numbers were paired to death numbers as $B10K=DK$ at the time of removal. So, we paired each removed ball to one of the balls that we put in: $B10\to D1,B20\to D2,B30\to D3,\dots$
, this means a ton of balls were never ever paired B1,B2,...,B9,B11,... etc.
The second thing that may happen with non random ball removal is also related to pairing at removal: we correlate BK=DK. We can do this by removing a ball with BK at each step K, which ensures that BK is paired to DK. This way each removed ball is paired with each ball that we put in, i.e. the same end result like in the random draw of removed balls. Obviously, this means that there are no balls left in the box after 12:00AM.
I just have shown that the problem has very little to do with probabilities per se. It has everything to do with powers of infinite countable (?) sets. The only real problem that I avoided discussing is whether the sets are truly countable. You see when you get closer to 12:00AM your rate of ball inserts is increasing rather quickly, to put it mildly. So, it's not so trivial to devise whether the number of balls that we put into the box is actually countable.
Unraveling
Now, I'm going to unravel this canonical solution of the paradox, and get back to our intuition.
How is is possible that we put 10 balls in, remove one and still run out of all the balls at 12 hour? Here's what really is happening. 12 hour is unreachable.
Let as reformulate the problem. We don't halve time intervals anymore. We put and remove balls every minute. Isn't this exactly the same as in the original problem? Yes and no. 
Yes, because nowhere in my exposition above I referred explicitly to time but at the very end. I was counting the steps k. So, we can keep counting the steps and dead balls by k.
No, because now we're never going to stop. We'll keep adding and removing balls till the end of time, which never arrives. While in the original problem the end is at 12 hour.
This explains how our intuition fails. Although we put balls at 9x rate of removal, because time never ends, every ball that we put in will be removed eventually! It may take infinite number of minutes, but it's Ok, because we have infinite number of minutes remaining. That's the true solution of the problem.
In this formulation would you ask "how many balls are in the box after infinity is over?" No! Because it's a nonsensical question. That's why the original question is nonsensical too. Or you could call it ill-posed.
Now, if you go back to the original problem, then the end of time apparently happens. It's at 12. The fact that we stopped putting balls in means that time just ended, and we reached beyond the end of it. So, the true answer to the question is that 12 o'clock should never occur. It's unreachable.
A: I want to make a reformulation that is as easy as possible to make the answer of 0 more intuitive, starting from the simplified example that balls are not removed randomly, but ball $n$ is removed at the $n$-th step.
Consider this: I put all balls into the urn at the beginning. In step 1, I take out ball 1. In step 2, I take out ball 2, and so on. Any doubt that the urn will be empty after infinite steps?
Okay. But if I don't put all balls into the urn at first, but only some balls, how could the urn be fuller in the end?
A: Let x be the number of balls that have been removed and y be the number of balls remaining. After each cycle y=9x. As x>0, y>0. There will be infinitely many balls in the urn at 12PM.
The reason that solutions based on probabilities lead to difficulties is that the probabilities from infinite series are tricky. ET Jaynes wrote about a few different apparent paradoxes of probability, like this one, in his book Probability Theory: The Logic of Science. I do not have my copy at hand, but the first part of the book is available online from Larry Bretthorst here. The following quote is from the preface.

Yet when all is said and done we find, to our own surprise, that
  little more than a loose philosophical agreement remains; on many
  technical issues we disagree strongly with de Finetti. It appears to
  us that his way of treating infinite sets has opened up a Pandora’s
  box of useless and unnecessary paradoxes; nonconglomerability and
  finite additivity are examples discussed in Chapter 15.
Infinite set paradoxing has become a morbid infection that is today
  spreading in a way that threatens the very life of probability theory,
  and requires immediate surgical removal. In our system, after this
  surgery, such paradoxes are avoided automatically; they cannot arise
  from correct application of our basic rules, because those rules admit
  only finite sets and infinite sets that arise as well-defined and
  well-behaved limits of finite sets. The paradoxing was caused by (1)
  jumping directly into an infinite set without specifying any limiting
  process to define its properties; and then (2) asking questions whose
  answers depend on how the limit was approached.
For example, the question: “What is the probability that an integer is
  even?” can have any answer we please in (0, 1), depending on what
  limiting process is to define the “set of all integers” (just as a
  conditionally convergent series can be made to converge to any number
  we please, depending on the order in which we arrange the terms).
In our view, an infinite set cannot be said to possess any “existence”
  and mathematical prop- erties at all—at least, in probability
  theory—until we have specified the limiting process that is to
  generate it from a finite set. In other words, we sail under the
  banner of Gauss, Kronecker, and Poincar ́e rather than Cantor,
  Hilbert, and Bourbaki. We hope that readers who are shocked by this
  will study the indictment of Bourbakism by the mathematician Morris
  Kline (1980), and then bear with us long enough to see the advantages
  of our approach. Examples appear in almost every Chapter.

The use of limits in the answer of @enumaris (+1) provides a way around the trickiness of infinities in probability.
A: It's worth reading amoeba's answer that is just excellent and clarifies the problem very much. I don't exactly disagree with his answer but want to point out that the solution of the problem is based on a certain convention. What is interesting is that this sort of problem shows that this convention, while often used, is questionable. 
Just as he says there is a technical point about proving that for each ball the probability to stay in the urn forever is 0. Apart from this point, the problem is not about probabilities. A deterministic equivalent may be given. It is much easier to understand. The key idea is: since every ball is absent from the urn from some point in time, the urn at the end is empty. If you represent the presence in the urn of each ball by a sequence of zeros and ones, each sequence is 0 from a certain range, thus its limit is 0.
Now the problem can be simplified even more. I call the moments 1, 2, 3.... for simplicity:


*

*moment 1: put ball 1 in the urn

*moment 2: remove it

*moment 3: put ball 2 in the urn

*moment 4: remove it

*moment 5: put ball 3 in the urn

*...


What balls at the end  (noon) ? With the same idea, same answer: none.
But fundamentally, there is no way to know, because the problem does not say  what happens at noon. Actually, it is possible that at the end of times, Pikachu comes suddenly in the urn. Or maybe balls all suddenly collapse and merge into one big ball. Not meaning that this is meant to be realistic, it's just not specified.
The problem can only be answered if a certain convention tells us how to go to the limit: a continuity assumption. The state of the urn at noon is the limit of its states before. Where should we look for a continuity assumption that would help us answer to the question?
In physical laws? Physical laws ensure a certain continuity. I think of a simplistic classical model, not calling on to real modern physics. But fundamentally, physical laws would bring exactly the same questions as the mathematical ones: the way we choose to describe continuity for physical laws relies on asking the question mathematically: what is continuous, how? 
We have to look for a continuity assumption in a more abstract way. The usual idea is to define the state of the urn as a function from the set of balls into $\{0;1\}$. 0 means absent, 1 means present. And to define continuity, we use product topology, aka pointwise convergence. We say that the state at noon, is the limit of the states before noon according to this topology. With this topology, there is a limit, and it is 0: an empty urn.
But now we modify the problem a little in order to challenge this topology:


*

*moment 1: put ball 1 in the urn

*moment 2: remove it

*moment 3: put ball 1 in the urn

*moment 4: remove it

*moment 5: put ball 1 in the urn

*...


For the same topology, the sequence of states has no limit. That's where I start to see the paradox as a true paradox. For me this modified problem is essentially the same. Imagine you are the urn. You see balls coming and going. If you can't read the number on it, whether it is the same ball or another one does not change what's happening to you. Instead of seeing balls as individual distinct elements, you see them as a quantity of matter coming in and out. The continuity could naturally be defined by looking at variations of the quantity of matter. And there is indeed no limit. In a way this problem is the same as the original problem where you decide to ignore the ball identity, thus leading to a different metric and a different notion of convergence. And even if you could see the number on the balls, the state could be seen as just a flickering presence with a growing number.
In one case, the limit of the sequence of your states is "empty", in the other case the limit is undefined.
The formalization of the problem with the product topology fundamentally relies on separating what happens to each different ball, and thus creating a metric reflecting the "distinguishablitiy". Only because of this separation, a limit can be defined. The fact that this separation is so fundamental to the answer but not fundamental for describing "what's going on" in the urn (a point that is endlessly arguable), makes me think the solution is the consequence of a convention rather than a fundamental truth.
For me, the problem, when considered as purely abstract has a solution as long as the missing information is provided: that the state at noon is the limit of the previous states and limit in what sense. However, when thinking of this problem intuitively, the limit of the sequence of states is not something you can think in a single manner. Fundamentally, I think there is no way to answer. 
A: Hurkyl (in an answer) and Dilip Sarwate (in a comment) give two common deterministic variants of this puzzle.  In both variants, at step $k$, balls $10k-9$ through $10k$ are added to the pile ($k=1,2,...$).  
In Hurkyl's variation, ball $k$ is removed.  In this variant, in can be definitively argued that there are no balls left because ball $n$ is removed at step $n$.
In Dilip Sarwate's variation, ball $10k$ is removed at step $k$, and so in this variant, all balls that are not multiples of $10$ remain.  In this variant, there are infinitely many balls in the urn at the end.
With these two variants as edge cases, we see that lots of different things can happen when doing this process.  For instance, you could arrange to have any finite set of balls remaining at the end, by doing Hurkyl's process but skipping the removal of certain balls.  In fact for any set $B$ with countably infinite complement (in the (positive) natural numbers), you can have that set of balls remaining at the end of the process.
We could look at the random variation of the problem (given in the original post) as selecting a function $f:\mathbb{N}\to\mathbb{N}$ with the conditions that (i) $f$ is one-to-one and (ii) $f(k)\le 10k$ for all $k\in \mathbb{N}$.
The argument given in the Sheldon Ross book (referenced in the post) shows that almost all (in the probabilistic sense) such functions are in fact onto functions (surjections).  
I see this as being somewhat analogous to the situation of selecting a number, $x$ from a uniform distribution on $[0,1]$ and asking what is the probability that the number is in the Cantor set (I am using the Cantor set rather than say the rational numbers because the Cantor set is uncountable).  The probability is $0$ even though there are many (uncountably many) numbers in the Cantor set that could have been chosen.  In the ball removing problem, the set of sequences in which there are any balls left is playing the role of the Cantor set.

Edit: BenMillwood correctly points out that there are some finite sets of balls that cannot be the remaining set. For instance, $1,2,...,10$ cannot be the remaining set.  You can have at most $90\%$ of the first $10n$ balls remaining for $n=1,2,3,...$. 
A: Enumaris' answer is perfectly right on the diverging limits problem. Nevertheless, the question can actually be answered in an unambiguous way. So, my answer will show you precisely where the zero balls solution goes wrong, and why the intuitive solution is the correct one.

It is true, that for any ball $n$, the probability of it being in the urn at the end $P(n)$ is zero. To be precise, it's only the limit that's zero: $P(n) = \lim_{N\to\infty} P(n, N) = 0$.
Now, you try to compute the sum
$$\lim_{N\to\infty} \operatorname{ballCount}(N) = \lim_{N\to\infty} \sum_{n=1}^{n\leq 10N} P(n,N).$$
The broken calculation jumps right in to that $P(n,N)$ part, saying that's zero in the limit, so the sum contains only terms of zero, so the sum is zero itself:
$$\begin{align}
\lim_{N\to\infty}\operatorname{ballCount}(N)
&= \lim_{N\to\infty} \sum_{n=1}^{n\leq 10N} P(n,N) \\
\text{broken step here }\longrightarrow
&= \lim_{N\to\infty} \sum_{n=1}^{n\leq 10N} \lim_{N\to\infty} P(n,N) \\
&= \lim_{N\to\infty} \sum_{n=1}^{n\leq 10N} P(n) \\
&= \lim_{N\to\infty} \sum_{n=1}^{n\leq 10N} 0 \\
&= \lim_{N\to\infty} 10 N\times 0 \\
&= 0
\end{align}$$
However, this is illegally splitting the $\lim$ into two independent parts. You cannot simply move the $\lim$ into the sum if the bounds of the sum depend on the parameter of the $\lim$. You must solve the $\lim$ as a whole.
Thus, the only valid way to solve this $\lim$ is to solve the sum first, using the fact that $\sum_{n=1}^{n\leq 10N} P(n,N) = 9N$ for any finite $N$.
$$\begin{align}
\lim_{N\to\infty} \operatorname{ballCount}(N)
&= \lim_{N\to\infty} \sum_{n=1}^{n\leq 10N} P(n,N) \\
&= \lim_{N\to\infty} 9N \\
&= \infty
\end{align}$$
The intuitive solution did precisely that, it's the "clever" solution that's fundamentally broken.
A: Ross describes three versions of this "paradox" in the Example 6a in his textbook. In each version, 10 balls are added to the urn and 1 ball is removed at each step of the procedure.


*

*In the first version, $10n$-th ball is removed at the $n$-th step. There are infinitely many balls left after midnight because all balls with numbers not ending in zero are still in there.

*In the second version, $n$-th ball is removed at the $n$-th step. There are zero balls left after midnight because each ball is eventually going to be removed at the corresponding step.

*In the third version, balls are removed uniformly at random. Ross computes the probability of each ball to be removed by step $n$ and finds that it converges to $1$ as $n\to\infty$ (note that this is not evident! one actually has to perform the computation). This means, by Boole's inequality, that the probability of having zero balls in the end is also $1$.
You are saying that this last conclusion is not intuitive and hard to explain; this is wonderfully supported by many confused answers and comments in this very thread. However, the conclusion of the second version is exactly as un-intuitive! And it has absolutely nothing to do with probability or statistics. I think that after one accepts the second version, there is nothing particularly surprising about the third version anymore.
So whereas the "probabilistic" discussion must be about the third version [see very insightful answers by @paw88789, @Paul, and @ekvall], the "philosophical" discussion should rather focus on the second version which is much easier and is similar in spirit to the Hilbert's hotel.

The second version is known as the Ross-Littlewood paradox. I link to the Wikipedia page, but the discussion there is horribly confusing and I do not recommend reading it at all. Instead, take a look at this MathOverflow thread from years ago. It is closed by now but contains several very perceptive answers. A short summary of the answers that I find most crucial is as follows.
We can define a set $S_n$ of the balls present in the urn after step $n$. We have that $S_1=\{2,\ldots 10\}$, $S_2=\{3,\ldots 20\}$, etc. There is a mathematically well-defined notion of the limit of a sequence of sets and one can rigorously prove that the limit of this sequence exists and is the empty set $\varnothing$. Indeed, what balls can be in the limit set? Only the ones that are never removed. But every ball is eventually removed. So the limit is empty. We can write $S_n \to \varnothing$.
At the same time, the number $|S_n|$ of the balls in the set $S_n$, also known as the cardinality of this set, is equal to $10n-n=9n$. The sequence $9n$ is obviously diverging, meaning that the cardinality converges to the cardinality of $\mathbb N$, also known as aleph-zero $\aleph_0$. So we can write that $|S_n|\to \aleph_0$.
The "paradox" now is that these two statements seem to contradict each other:
\begin{align}
S_n &\to \varnothing \\
|S_n| &\to \aleph_0 \ne 0
\end{align}
But of course there is no real paradox and no contradiction. Nobody said that taking cardinality is a "continuous" operation on sets, so we cannot exchange it with the limit:$$\lim |S_n| \ne |\lim S_n|.$$ In other words, from the fact that $|S_n|=9n$ for all integer $n\in \mathbb N$ we cannot conclude that $|S_\omega|$ (the value at the first ordinal) is equal to $\infty$. Instead, $|S_\omega|$ has to be computed directly and turns out to be zero. 

So I think what we get out of this really is the conclusion that taking cardinalities is a discontinous operation... [@HarryAltman]
So I think this paradox is just the human tendency to assume that "simple" operations are continuous. [@NateEldredge]


This is easier to understand with functions instead of sets. Consider a characteristic (aka indicator) function $f_n(x)$ of set $S_n$ which is defined to be equal to one on the $[n, 10n]$ interval and zero elsewhere. The first ten functions look like that (compare the ASCII art from @Hurkyl's answer):
$\quad\quad\quad$
Everybody will agree that for each point $a\in\mathbb R$, we have $\lim f_n(a) = 0$. This by definition means that functions $f_n(x)$ converge to the function $g(x)=0$. Again, everybody will agree to that. However, observe that the integrals of these functions $\int_0^\infty f(x)dx = 9n$ get larger and larger and the sequence of integrals diverges. In other words,
$$\lim\int f_n(x)dx \ne \int \lim f_n(x) dx.$$
This is a completely standard and familiar analysis result. But it is an exact reformulation of our paradox!

A good way to formalize the problem is to describe the state of the jug not as a set (a subset of $\mathbb N$), because those are hard to take limits of, but as its characteristic function. The first "paradox" is that pointwise limits are not the same as uniform limits. [@TheoJohnson-Freyd]

The crucial point is that "at midnight noon" the whole infinite sequence has already passed, i.e. we made a "trasfinite jump" and arrived to the transfinite state $f_\omega = \lim f_n(x)$. The value of the integral "at midnight noon" has to be the value of the integral of $\lim f_n$, not the other way around.

Please note that some of the answers in this thread are misleading despite being highly upvoted.
In particular, @cmaster computes $\lim_{n\to\infty} \operatorname{ballCount}(S_n)$ which is indeed infinite, but this is not what the paradox asks about. The paradox asks about what happens after the whole infinite sequence of steps; this is a transfinite construction and so we need to be computing $\operatorname{ballCount}(S_\omega)$ which is equal to zero as explained above.
A: This argument is focused on the tendency for infinite sets and sequences to behave in unitnuitive ways.  This is no more surprising than the Hilbert Hotel.  In such a case, you will indeed have taken out an infinite number of balls, but you will have put an infinite number in.  Consider the Hilbert Hotel in reverse.  You can remove an infinite number of guests from the hotel, and still have an infinite number left.
Whether this is physically realizable is another question entirely.
As such, I would consider it not necessarily ill formed, but rather put in the wrong book.  This sort of counting question belongs in a set theory course, not a probability course.
A: I think it helps to remove the superfluous temporal component of the problem.
The more basic variant of this paradox is to always remove the lowest numbered ball. For ease of drawing, I will also only add two balls at each step.
The procedure describes how to fill out an infinite two-dimensional grid:
.*........
..**......
...***....  ....
....****..
.....*****

 :  :  :
 :  :  :

where each row is formed from the previous by adding two asterisks on the right then removing the leftmost.
The questions one then asks are:

How many columns end with repeated asterisks rather than repeated dots?

In my opinion, the idea to mistakenly equate this result with "the limit of the number of asterisks in each row" is much less compelling.
A: This answer aims to do four things: 


*

*Review Ross's mathematical formulation of the problem, showing how it follows directly and unambiguously from the problem description. 

*Defend the position that Ross's paradoxical solution is both mathematically sound and relevant to our understanding of the physical world, whether or not it is 100% physically realizable.

*Discuss certain fallacious arguments rooted in physical intuition, and show that the oft-stated "physical" solution of infinite balls at noon is not only in contradiction to mathematics, but to physics as well.

*Describe a physical implementation of the problem which may make Ross's solution more intuitive. Start here for the answer to Carlos's original question.
1. How to Describe the Problem Mathematically
We will unpack the initial "infinite process modeling" step of Ross's argument (p. 46). Here is the statement we will focus on justifying:

Define $E_n$ to be the event that ball number 1 is still in
  the urn after the first n withdrawals have been made... The event that ball number 1 is in the urn at 12 P.M. is just the event $\bigcap_{n=1}^\infty E_n$.

Before we unpack Ross's statement, let's consider how it is even possible to understand the urn's contents at noon, after an infinite sequence of operations. How could we possibly know what is in the urn? Well, let's think about a specific ball $b$; you can imagine $b=1$ or $1000$ or whatever you want. If ball $b$ was taken out at some stage of the process before noon, certainly it won't be in the urn at noon. And conversely, if a given ball was in the urn at every single stage of the process up until noon (after it was added), then it was in the urn at noon. Let's write these statements out formally:

A ball $b$ is in the urn at noon if and only if it was in the urn at every stage $n \in \{n_b, n_b + 1, n_b + 2, ...\}$ before noon, where $n_b$ is the stage the ball was added to the urn.

Now let's unpack Ross's statement - what does $\bigcap_{n=1}^\infty E_n$ mean in plain English? Let's take a single realization $x$ of the urn process and talk it out:


*

*$x \in E_1$ means that ball 1 is in the urn after stage 1 of the process.

*$x \in E_1 \bigcap E_2$ means that ball 1 is in the urn after stages 1 and 2 of the process.

*$x \in E_1 \bigcap E_2 \bigcap E_3$ means that ball 1 is in the urn after stages 1, 2, and 3 of the process.

*For any $k \in \{1, 2, 3, ...\}$, $x \in \bigcap_{k=1}^n E_k$ means that the ball is in the urn after stages $1$ thru $n$.


Clearly, then, $x \in \bigcap_{k \in \{1, 2, 3...\}} E_k$ means that, in realization $x$ of this urn process, ball 1 is in the urn after stages 1, 2, 3, et cetera: all finite stages $k$ before noon. The infinite intersection $\bigcap_{n = 1}^\infty E_n$ is just another way of writing that, so $\bigcap_{n = 1}^\infty E_n$ contains precisely the realizations of the process where ball 1 was in the urn at all stages before noon. An event is just a defined set of realizations of a process, so the last sentence is precisely equivalent to saying that $\bigcap_{n = 1}^\infty E_n$ is the event that ball 1 was in the urn at all stages before noon, for this random process.
Now, the punchline: by our "if and only if" statement above, this is exactly the same as saying that ball 1 was in the urn at noon! So $\bigcap_{n = 1}^\infty E_n$ is the event that ball 1 is in the urn at noon, just as Ross originally stated. QED
In the derivation above, everything we said is equally valid for both the deterministic and probabilistic versions, because deterministic modeling is a special case of probabilistic modeling in which the sample space has one element. No measure theoretic or probability concepts were even used, beyond the words "event" and "realization" (which are just jargon for "set" and "element"). 
2. The Paradoxical Solution is Mathematically Sound and Relevant to Physics
After this setup point, the deterministic and probabilistic variants diverge. In the deterministic variant (version 2 from amoeba's post), we know ball 1 is taken out on the first step, so $E_1 = \emptyset$ and the infinite intersection, of course, is also empty. Similarly, any other ball $b$ is taken out at stage $b$ and is not present at noon. Thus the urn cannot contain any numbered ball $b$ at noon and must therefore be empty.
In the probabilistic variant, the same phenomenon happens, just in a softer "in-expectation" sense. The probability of any given ball's being present dwindles to zero as we approach noon, and at the limiting time of noon, the ball is almost surely not present. Since each ball is present with probability zero, and the sum of infinitely many zeros is still zero, there are almost surely no balls in the urn at noon. All of this is shown completely rigorously by Ross; details can be filled in with a knowledge of graduate-level measure theory, as @ekvall's answer shows.
If you accept the standard arguments about mathematical objects expressed as infinite sequences, for example $0.999... = 1$, the argument here should be just as acceptable, as it relies on the exact same principles. The only question remaining is whether the mathematical solution applies to the real world, or just the platonic world of mathematics. This question is complex and is discussed further in section 4.
That said, there is no reason to presuppose that the infinite urn problem is unphysical, or to reject it as irrelevant even if it is unphysical. Many physical insights have been gained from studying infinite structures and processes, for example, infinite wires and percolation lattices. Not all of these systems are necessarily physically realizable, but their theory shapes the rest of physics. Calculus itself is "unphysical" in some ways, because we don't know if it is possible to physically realize the arbitrarily small distances and times that are often its subject of study. That doesn't stop us from putting calculus to incredibly good use in the theoretical and applied sciences.
3. The Unphysicality of Solutions Based on "Physical Intuition"
For those who still believe that Ross's math is wrong or physically inaccurate in the deterministic variant, and the true physical solution is infinitely many balls: regardless of what you think happens at noon, it is impossible to deny the situation before noon: every numbered ball added to the urn eventually gets removed. So if you think there are somehow still infinitely many balls in the urn at noon, you must admit that not one of those balls can be a ball added before noon. So those balls must have come from somewhere else: you are asserting that infinitely many balls, unrelated to the original problem process, suddenly pop into existence at precisely noon to rescue the continuity of cardinality from being violated. As unphysical as the "empty set" solution might seem intuitively, this alternative is objectively and demonstrably unphysical. Infinite collections of objects do not pop into being in an instant just to satisfy poor human intuitions about infinity.
The common fallacy here seems to be that we can just look at the number of balls as time approaches noon, and assume that the divergent trend yields infinitely many balls at noon, without regard to exactly which balls are being taken in and out. There has even been an attempt to justify this with the "principle of indifference", which states that the answer shouldn't depend on whether the balls are labeled or not.
Indeed, the answer does not depend on whether the balls are labeled or not, but that is an argument for Ross's solution, not against it. From the perspective of classical physics, the balls are effectively labeled whether you think of them as labeled or not. They have distinct, permanent identities which are equivalent to labels, and a truly physical analysis must account for this, whether or not numbers are literally written on the balls. The labels themselves do not directly affect how the solution comes out, but they are needed to describe exactly how the balls are moved around. Some procedures leave balls in the urn forever, others provably remove every ball that is added, and labels are needed to even describe the difference between these procedures. Attempting to ignore the labels is not "physical", it's just neglecting to understand the physical problem precisely enough to solve it. (The same goes for complicated variants that reshuffle the labels at each stage. What matters is which balls are in the urn, not the labels someone has placed or replaced on them. This can be determined by ignoring the complicated relabeling scheme entirely and simply using a single unchanging labeling scheme, the one of Ross's original problem.)
The only way distinguishability would fail to be true is if the "balls" were quantum mechanical particles. In this case, the indifference principle fails spectacularly. Quantum physics tells us that indistinguishable particles behave completely differently than distinguishable ones. This has incredibly fundamental consequences for the structure of our universe, such as the Pauli exclusion principle, which is perhaps the single most important principle of chemistry. No one has attempted to analyze a quantum version of this paradox yet.
4. Describing the Solution Physically
We have seen how vague "physical" intuitions can lead us astray on this problem. Conversely, it turns out that a more physically precise description of the problem helps us understand why the mathematical solution is actually the one that makes the most physical sense.
Consider an infinite Newtonian Universe governed by the laws of classical mechanics. This Universe contains two objects: an infinite Shelf and an infinite Urn, which start at the Origin of the Universe and run alongside one another, one feet apart, forever and ever. The Shelf lies on the line $y = 0$ feet, while the Urn lies on the line $y = 1$ feet. Along the Shelf are laid infinitely many identical balls, evenly spaced one foot apart, the first being one foot from the Origin (so ball $n$ is on the line $x = n$ feet). The Urn - which is really just like the Shelf, but a bit more ornate, closed over, and generally Urnish - is empty. 
An Aisle connects the Shelf and Urn at the bottom, and on top of the Aisle, at the Origin, sits an Endeavor robot with an infinite power supply. Beginning at 11 AM, Endeavor activates and begins zooming back and forth in the Aisle, transferring balls between Urn and Shelf according to Ross-Littlewood's programmed instructions:


*

*When the program commands ball $n$ to be inserted into the Urn, the ball $n$ feet from the Origin is transferred from the Shelf to the Urn.

*When the program commands ball $n$ to be removed from the Urn, the ball $n$ feet from the Origin is transferred from the Urn to the Shelf.


In either case, the transfer is made straight across, so the ball remains $n$ feet from the Origin. The process unfolds as specified in the Ross-Littlewood problem:


*

*At 11:00 AM, Endeavor transfers balls 1-10 from Shelf to Urn, then moves one of the Urn balls back to Shelf. 

*At 11:30 AM, Endeavor transfers balls 11-20 from Shelf to Urn, then moves one of the Urn balls back to Shelf.

*At 11:45 AM, Endeavor transfers balls 21-30 from Shelf to Urn, then moves one of the Urn balls back to Shelf.

*et cetera...
As the process continues, each new step requires longer trips up and down the Aisle, and only half the time to make the trips. Thus, Endeavor must move up and down the Aisle exponentially faster as noon closes in. But it always keeps up with the program, because it has an infinite power supply and can move as fast as needed. Eventually, noon arrives.
What happens in this more vividly imagined version of the paradox? Watched from above, the approach towards noon is truly spectacular. Within the Urn, a Wave of balls appears to propagate outward from the Origin. The Wave's size and speed grow without bound as noon approaches. If we were to take pictures immediately after each step what would the layout of balls would look like? In the deterministic case, they would look exactly like the step functions in amoeba's answer. The ball positions $(x, y)$ would follow precisely the curves he has plotted. In the probabilistic case, it would look roughly similar, but with more straggling near the Origin.
When noon arrives, we take stock of what has happened. In the deterministic version, each ball was transferred from the Shelf to the Urn exactly once, then moved back at a later step, with both transfers happening before noon. At noon, the Universe must be back to its original 11 AM state. The Wave is no more. Each ball is back exactly where it started. Nothing has changed. The Urn is empty. In the probabilistic version the same thing happens, except now the result is only almost sure rather than sure. 
In either case, "physical objections" and complaints about infinity seem to vanish into thin air. Of course the Urn is empty at noon. How could we have imagined otherwise?
The only remaining mystery is the fate of Endeavor. Its displacement from the Origin and its velocity became arbitrarily large as noon approached, so at noon, Endeavor is nowhere to be found in our infinite Newtonian Universe. The loss of Endeavor is the only violation of physics which has occurred during the process.
At this point, one could object that Endeavor is not physically possible, since its speed grows without bound and would eventually violate the relativistic limit, the speed of light. However, we can change the scenario slightly to resolve this issue. Instead of a single robot, we could have infinitely many robots, each responsible for a single ball. We could program them beforehand to ensure perfect coordination and timing according to Ross's instructions.
Is this variation 100% physical? Probably not, because the robots would have to operate with arbitrarily precise timing. As we approach noon, the precision demanded would eventually fall below the Planck time and create quantum mechanical issues. But ultimately, an infinite wire and an infinite percolation lattice might not be all that physical either. That doesn't stop us from studying infinite systems and processes and determining what would happen if the obstructing physical constraints were suspended.
4a. Why Count Monotonicity is Violated
A number of Ross skeptics have questioned how it is possible that the number of balls in the urn increases without bound as we approach noon, then is zero at noon. Ultimately we must believe in rigorous analysis over our own intuition, which is often wrong, but there is a variation of the paradox that helps illuminate this mystery.
Suppose that instead of infinitely many balls, we have $10N$ balls labeled 1, 2, 3, up to $10N$, and we issue the following addition to the rules for the ball mover:


*

*If the instructions ask you to move a ball that does not exist, ignore that instruction.


Note that the original problem is unchanged if we add to it this instruction, since the instruction will never be activated with infinitely many balls. Thus, we can think of the original problem and this new family of problems to be part of the same family, with the same rules. Examining the finite $N$ family, especially for very large $N$, can help us to understand the "N = $\infty$" case.
In this variation, the balls accumulate 9 per step as before, but only up to step $N$ of the process. Then the numbers for balls to be added no longer correspond to actual balls, and we can only comply with the instruction to remove balls, and the process stops after $9N$ additional steps, for a total of $10N$ steps. If $N$ is very large, the removal-only phase occurs very close to noon, when the tasks are being done very rapidly, and the urn is emptied out very quickly. 
Now suppose we do this variation of the experiment for each value of $N$ and graph the ball count over time, $f_N(t)$, where $t$ ranges from 0 to 1 hour after 11AM (i.e. 11AM to noon). Typically $f_N(t)$ rises for a while, then falls back to zero at or before $t=1$. In the limit as $N$ approaches infinity, the graph rises ever higher and the fall is ever more rapid. By noon the urn is always empty: $f_N(1) = 0$. In the limiting graph, $f(t) = \lim_{N \rightarrow \infty} f_N(t)$, the curve approaches infinity for $t < 1$ but $f(1) = 0$. This is precisely the result derived in Ross's proof: the ball count diverges to infinity before noon, but is zero at noon. In other words, Ross's solution preserves continuity with respect to N: the pointwise limit of the ball count as $N \rightarrow \infty$ matches the ball count in the infinite ball case.
I do not consider this a primary argument for Ross's solution, but it may be helpful for those who are puzzled about why the ball count goes up forever, than crashes to zero at noon. While strange, it is the limiting behavior of the finite version of the problem as $N \rightarrow \infty$, and thus does not come as a "sudden shock" in the infinite case.
A Final Reflection
Why has this problem proven to be such a tar-pit for so many? My speculation is that our physical intuition is much vaguer than we think it is, and we often draw conclusions based on imprecise and incomplete mental conceptions. For example, if I ask you to think of a square that is also a circle, you may imagine something squarish and circlish, but it won't be precisely both of those things - that would be impossible. The human mind can easily mash together vague, contradictory concepts into a single mental picture. If the concepts are less familiar, like the Infinite, we can convince ourselves that these vague mental mashups are actually conceptions of the Real Thing.
This is precisely what happens in the urn problem. We do not really conceive of the whole thing at once; we think about bits and pieces of it, like how many balls there are over time. We wave away supposedly irrelevant technicalities, like what happens to each humble little ball over time, or how exactly an "urn" can hold infinitely many balls. We neglect to set out all the details precisely, not realizing that the result is a mashup of inconsistent, incompatible mental models.
Mathematics is designed to rescue us from this condition. It disciplines and steels us in the face of the unfamiliar and the exotic. It demands that we think twice about the things that "must" be true... right? It reminds us that no matter how strange things get, one and one is still two, a ball is either in an urn or it is not, and a statement is either true or false. If we persevere, these principles eventually bring clarity to most of our problems.
Those who subordinate mathematical analysis to "physical" or "common-sense" intuitions do so at their peril. Hand-waving about intuitions is only the start of physics. Historically, all successful branches of physics have eventually founded themselves on rigorous mathematics, which culls away incorrect physical intuitions, strengthens correct ones, and enables the rigorous study of ideal systems, such as the infinite current-carrying wire, which illuminate the behavior of the more complicated, messy real world. Ross-Littlewood is a physical problem, typically interpreted as one of classical mechanics, and classical mechanics has a completely mature and rigorous mathematical foundation. We should rely upon mathematical modeling and analysis for our intuitions about the world of classical physics, not the other way around.
A: Several posters have been concerned the computations in Ross may not be rigorous. This answer addresses that by proving the existence of a probability space where all sets of outcomes considered by Ross are indeed measurable, and then repeats the vital parts of Ross's computations.
Finding a suitable probability space
To make Ross's conclusion that there are no balls in the urn at 12 P.M., almost surely, rigorous, we need the existence of a probability space $(\Omega, \mathcal F, P)$ where the event "no balls in the urn at 12 P.M." can be constructed formally and shown to be measurable. To that end, we shall use Theorem 33 [Ionescu - Tulcea] in these lecture notes, slightly reworded, and a construction suggested by @NateEldredge in a comment to the question. 
Theorem. (Ionescu - Tulcea Extension Theorem) Consider a sequence
of measurable spaces $(\Xi_n, \mathcal X_n), n = 1, 2, \dots$. Suppose that for each $n$, there exists a probability kernel $\kappa_n$ from $(\Xi_1, \mathcal X_1) \times \dots \times(\Xi_{n-1}, \mathcal X_{n-1})$ to $(\Xi_n, \mathcal X_n)$ (taking $\kappa_1$ to be a kernel insensitive to its first argument, i.e., a probability measure). Then there exists a sequence of random variables $X_n, n = 1, 2, \dots$ taking values in the corresponding $\Xi_n$, such that, for every $n$, the joint distribution of $(X_1, \dots, X_n)$ is that implied by the kernels $\kappa_1, \dots, \kappa_n$.
We let $X_n$ denote the label of the ball removed at the $n$th withdrawal. It's clear that the (infinite) process $X = (X_1, X_2, \dots)$, if it exists, tells us everything we need to know to mimic Ross's arguments. For example, knowing $X_1, \dots, X_m$ for some integer $m \geq 0$ is the same as knowing the number of balls in the urn after withdrawal $m$: they are precisely the added balls with labels $\{1, 2, \dots, 10m\}$, minus the removed balls $\{X_1, \dots, X_m\}$. More generally, events describing which, and how many, balls are in the urn after any given withdrawal can be stated in terms of the process $X$.
To conform with Ross's experiment we need that, for every $n\geq 2$, the distribution of  $X_n \mid X_{n-1}, \dots, X_{1}$ is uniform on $\{1, 2, \dots, 10n\} \setminus {X_1, \dots, X_{n-1}}$. We also need the distribution of $X_1$ to be uniform on $\{1,\dots, 10\}$. To prove that an infinite process $X = (X_1, X_2, \dots)$ with these finite-dimensional distributions indeed exists, we check the conditions of the Ionescu-Tulcea Extension Theorem. For any integer $n$, let $\mathcal I_n = \{1, 2, \dots, n\}$ and define the measurable spaces $(\Xi_n, \mathcal X_n) = (\mathcal I_{10n}, 2^{\mathcal I_{10n}})$, where $2^B$ denotes the power set of the set $B$. Define the measure $\kappa_1$ on $(\Xi_1, \mathcal X_1)$ to be the one that puts mass $1/10$ on all elements of $\Xi_1$. For any $n  \geq 2$, and $(x_1, \dots, x_{n-1}) \in \Xi_1 \times \dots \times \Xi_{n-1}$ define $\kappa_n(x_1, \dots, x_{n-1}, \cdot)$ to be the probability kernel that puts equal mass on all points in $\Xi_n \setminus \{x_1, \dots, x_{n-1}\}$, and mass zero on all other points, i.e. on the integers $x_i \in \Xi_n, i = 1, \dots, n - 1$. By construction, the probability kernels agree with the uniform removal probability specified by Ross. Thus, the infinite process $X$ and the probability space $(\Omega, \mathcal F, P)$, the existence of which are given by the theorem, give us a way to formally carry out Ross's argument.
Let $E_{in}$ denote the set of outcomes such that ball $i$ is in the urn after withdrawal $n$. In terms of our stochastic process $X$ this means that, for all $i$ and $n$ such that $i \leq 10n$ we define $E_{in} = \cap_{j = 1}^n\{\omega: X_j(\omega) \neq i\}$, i.e. ball $i$ was not removed in any of the draws up to and including the $n$th. For $i > 10n$ we can clearly define $E_{in} = \emptyset$ since ball $i$ has not yet been added to the turn. For every $j$ and $i$, the set $\{\omega: X_j(\omega) \neq i\}$ is measurable since $X_j$ is a random variable (measurable). Thus, $E_{in}$ is measurable as the finite interesection of measurable sets. 
We are interested in the set of outcomes such that there are no balls in the urn at 12 P.M. That is, the set of outcomes such that for every integer $i = 1, 2\dots$, ball $i$ is not in the urn at 12 P.M. For every $i$, let $E_i$ be the set of outcomes ($\omega \in \Omega$) such that ball $i$ is in the urn at 12 P.M. We can construct $E_i$ formally using our $E_{in}$ as follows. That $i$ is in the urn at 12 P.M. is equivalent to it being in the urn after every withdrawal made after it was added to the urn, so $E_i = \cap_{n:i\leq 10n}E_{in}$. The set of outcomes $E_i$ is now measurable as the countable intersection of measurable sets, for every $i$. 
The outcomes for which there is at least one ball in the urn at 12 P.M. are those for which at least one of the $E_i$ happen, i.e. $E = \cup_{i = 1}^\infty E_i$. The set of outcomes $E$ is measurable as the countable union of measurable sets. Now, $\Omega \setminus E$ is the event that there are no balls in the urn at 12 P.M., which is indeed measurable as the complement of a measurable set. We conclude that all desired sets of outcomes are measurable and we can move on to computing their probabilities, as Ross does.
Computing the probability $P(\Omega \setminus E)$
We first note since the family of events $E_i, i = 1, 2, \dots$ is countable, we have by countable sub-additivity of measures that 
$$
P(E) \leq \sum_{i = 1}^\infty P(E_i) = \lim_{N\to \infty}\sum_{i = 1}^N P(E_i).
$$
For ease of notation, let's denote the real number $P(E_i) = a_i$ for all $i$. Clearly, to show that $P(E) = 0$ it suffices to show that $\sum_{i = 1}^N a_i = 0$ for all $N$. This is equivalent to showing that $a_i = 0$ for every $i$, which we shall do now.
To that end, note that for all $n$ such that ball $i$ has been added to the urn, i.e. $10n \geq i$,  $E_{in} \supseteq E_{i(n + 1)}$. This is so because if ball $i$ is in the urn at step $n + 1$, it is also in the urn at step $n$. In other words, the sets $E_{in}$, form a decreasing sequence for all $n$ such that $10n \geq i$. For ease of notation, let $a_{in} = P(E_{in})$. Ross proves that $a_{1n} \to 0$ as $n \to \infty$ and states that this can also be shown for all other $i$, which I will take as true. The proof consists of showing that $a_{in} = \prod_{k = i}^n[9k / (9k + 1)]$ and $\lim_{n \to \infty}a_{in} = 0$ for all $i$, an elementary but lenghty calculation I will not repeat here. Armed with this result, and the fact that the family of events $E_{in}$, $10n > i$ is countable for every i, continuity of measures gives
$$a_i = P(\cap_{n: 10n > i}E_{in}) = \lim_{n \to \infty} P(E_{in}) = \lim_{n \to \infty}a_{in} = 0.$$ 
We conclude that $P(E) = 0$, and thus $P(\Omega\setminus E) = 1$ as claimed. QED.

Some common misunderstandings:


*

*One answer is concerned with the fact that (in my notation) $\lim_{N \to \infty}\sum_{i = 1}^N \lim_{n \to \infty }a_{in} \neq \lim_{N \to \infty}\sum_{i = 1}^N a_{iN}$. This, however, has no bearing on the validity of the solution bevause the quantity on the right hand side is not the one of interest per the provided argument.

*There has been some concern that the limit cannot be moved inside the sum, or in other words cannot be interchanged with the sum in the sense that it may be the case that $\sum_{i = 1}^\infty \lim_{n \to \infty}a_{in} \neq \lim_{n \to \infty}\sum_{i = 1}^\infty a_{in}$. Like the previous remark, this is irrelevant to the solution because the quantity on the right hand side is not the one of interest.

A: 
On the one hand, you could try to explain it like this: "think of the
  probability of any ball i being on the urn at 12 P.M. During the
  infinite random draws, it will eventually be removed. Since this holds
  for all balls, none of them can be there at the end".

I don't find this argument convincing.  If this argument works, then the following argument works:  Every year, some people are born (say a constant fraction of the total population), and some people die (suppose a constant fraction).  Then, since in the limit any particular person is almost surely dead, then the human race must go extinct!  Now, the human race may go extinct for other reasons, but this argument is garbage.
It doesn't make any sense for this problem to have one solution when the balls are numbered and for it to have a totally different answer when the balls are anonymous.  By symmetry, arbitrary labels should not affect the solution.  Jaynes called this argument the principle of indifference, which I accept.
In other words, if someone told you that they put ten balls into an urn and remove one repeatedly, and how full is the urn in the limit, would your answer be "It depends on whether the balls are numbered"?  Of course not.  That urn's contents diverge just like the urn in this problem.
Therefore, I think the solution lies in how we formalize the problem.  From the usual definition of set-theoretic limit, we have
$$\liminf_{n \to \infty} S_n = \bigcup_{n \ge 1} \bigcap_{j \geq n} S_j.$$
$$\limsup_{n \to \infty} S_n = \bigcap_{n \ge 1} \bigcup_{j \geq n} S_j$$
Let the limit of the cardinality of the set be
$$k\triangleq \lim_{n\to\infty}|S_n|$$
and the cardinality of the $\liminf$-limit of the set be
$$l \triangleq \left|\liminf_{n\to\infty} (S_n)\right|.$$
I propose that set-theoretic limit be redefined so that:
\begin{align}
  \lim_{n\to\infty} S_n &\triangleq
  \begin{cases}
    \liminf_{n\to\infty} (S_n) &\text{if } \liminf_{n\to\infty} (S_n) = \limsup_{n\to\infty} (S_n), k \text{ exists, and }k=l \\
    \alpha_k &\text{if }\liminf_{n\to\infty} (S_n) = \limsup_{n\to\infty} (S_n), k\text{ exists, and }k \ne l \\
    \text{undefined} &\text{otherwise.}
  \end{cases}
\end{align}
This special “anonymous set” $\alpha_k$ describes what happens at infinity.  Just as $\infty$ stands in for the limiting behavior of numbers, $\alpha$ stands in for the limiting behavior of sets.  Namely, we have $i \notin \alpha_k \forall i$, and $|\alpha_k| = k$.  The benefit of this formalism is that it gives us continuity of cardinality and consistency with the principle of indifference.
For the urn problem, we have $S_n = \{n+1, \dotsc, 10n\}$ is the set of balls in the urn.  And $$\lim_{n\to\infty} S_n = \alpha_{\infty}.$$
Thus, the elements don't "fall off a cliff" at infinity, which doesn't make sense any more than it makes sense for humanity to go extinct merely because no man is immortal.
Similarly, suppose we modify the problem so that at each step one ball is added and the lowest-numbered ball is removed.  Then, how many balls are in the urn in the limit?  Anonymous sets give the intuitive answer:
$$\lim_{n\to\infty}\{n\} = \alpha_1.$$
I recognize that mathematicians can disagree about resolutions to this paradox, but to me, this is the most intuitive resolution.
A: OK, I'll try again.
The answer is that the paradox is purely mathematical. Enumaris's and cmaster's answer's tell what is going on in one way, but this is another way to see the problem. The problem is how we deal with probabilities with infinities, as Jaynes has written about (see my other attempted answer for details). 
An infinite series is usually treated as if it has no end, but in this problem there is an end time (12PM) and so logically, even if not mathematically, there is a last cycle of addition and removal of balls: the one that happens infinitessimally prior to 12PM. The existence of a 'last' cycle allows us to look at the probabilities backwards as well as forwards through time.
Consider the ten balls last added. For each of them their probability of being removed is zero because they are each just one of infinity balls that might be removed. Thus the probability that there will be at least ten balls remaining at 12PM is unity.
QED. A probabilistic argument that does not lead to nonsense.
A: I believe that this example supports "if the premise is false then the conditional is true"
In this universe, there are no infinite urns and no infinite collection of balls.  It is impossible to split time into arbitrarily small pieces.
Thus Sheldon Ross is right to say that the urn is empty at 12:00.  Students who say that the urn has infinite balls at 12:00 are just as right.
If you answered the urn has 50 balls then you are also correct.
I have not rigorously proved that this universe does not contain infinite urns and infinite balls and that time is not atomic - I just believe those things.  If you believe those three assertions are wrong, then you believe Ross's problem is empirically falsifiable.  I am waiting for your experimental results.
A: I support the opinion that the problem is ill-posed. When we consider something transfinite we often have to use a limit. It seems that here it is the only way. Since we distinguish different balls, we have an infinite-dimensional process  $$(X_{t,1}, X_{t,2},...),$$ 
where $t=-1,-1/2,-1/4,...$ stands for the time, $X_{t,j}=1$ if there is the ball $j$ at time $t+0$ and $X_{t,j}=0$ otherwise. 
Now it is on each everyone's discretion which convergence to use: uniform, componentwise, $l_p$, etc. Needless to say, the answer depends on the choice.
The misunderstanding in this problem goes from neglecting the fact that metric issues are crucial when we consider convergence of infinite-dimensional vectors. Without choosing the type of convergence, no correct answer can be given.
(There is componentwise convergence to zero vector. While $l_1$ norm counts the number of balls, so in this norm the process is exploding.)
A: Recently several comments by Wilhelm, Wolfgang Mückenheim, caused me to reconsider certain formulations in my answer. I am posting this as a new answer mainly because the different approach of this answer, not arguing about the teaching of this problem, but instead about the paradox being invalid.
Wilhelm discusses in his lengthy manuscript that 

Transactions are only possible
  at finite steps $n$ (there is no action possible "between all $n$ and $\omega$").

This reminded me of the term 
$$\sum_{k=1}^\infty \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right)$$
which is derived from Ross' work. This term is indeterminate when the path to infinity is not defined for the following limit.
$$\lim_\limits{(l,m)\to(\infty,\infty)}\sum_{k=1}^l\prod_{n=k}^m\left(\frac{9n}{9n+1}\right)$$
This seems to resemble the point that Wilhelm discusses and is also mentioned in aksakal's answer. The steps in time become infinitely small, so we will be able to reach 12 p.m. in that sense, but we will at the same time need to add and remove an (unphysical) infinity number of balls. It is a false idea to attach this supertask to a process like Zeno's arrow, just like the switch of Thompson's paradoxical lamp can not have a definite position at the end of a supertask.
In terms of the limit we can say that the physical path to infinity that we take is 
$$\lim_\limits{l\to \infty}\sum_{k=1}^l\prod_{n=k}^l\left(\frac{9n}{9n+1}\right) = \lim_\limits{l\to \infty} \frac{9l}{10}$$
so not zero but infinite.
A: The problem as stated is a variant of a conditionally convergent sum.  That is, the sum is indeterminate depending on how the addition is performed; in what order the terms are summed. In general, if $\Sigma_{n=0}^\infty x_n$ converges but  $\Sigma_{n=0}^\infty |x_n|$ is divergent, then convergence is conditional and not absolute. Although this is most often applied to alternating sign series, the order of signs is either irrelevant or the series is conditionally convergent. As such, there is no unique answer, see Mathworld commentary.
Finally, the only paradox inherent in finding different answers for a conditionally convergent sum is how one can expect it to be otherwise. This is worse than asking a pseudo-random number generator to create equal valued answers for different seeds because in that latter case one would not typically obtain as well $\pm\infty$ as admissible answers.
A: More intuition than formal education, but:
If the intervals to midnight are halving, we never reach midnight... we only approach asymptotically; so one could argue that there is no solution.
Alternatively, depending on the phrasing:


*

*as there are infinite intervals of +10 balls the answer is infinite

*as there are infinite intervals of (+10 balls - 1) the answer is 10*infinite -1*infinite = 0?

*as there are infinite intervals of (+9 balls) +1 the answer is infinite + 1

