What happens if the explanatory and response variables are sorted independently before regression? Suppose we have data set $(X_i,Y_i)$ with $n$ points. We want to perform a linear regression, but first we sort the $X_i$ values and the $Y_i$ values independently of each other, forming data set $(X_i,Y_j)$. Is there any meaningful interpretation of the regression on the new data set? Does this have a name?
I imagine this is a silly question so I apologize, I'm not formally trained in statistics. In my mind this completely destroys our data and the regression is meaningless. But my manager says he gets "better regressions most of the time" when he does this (here "better" means more predictive). I have a feeling he is deceiving himself.
EDIT: Thank you for all of your nice and patient examples. I showed him the examples by @RUser4512 and @gung and he remains staunch. He's becoming irritated and I'm becoming exhausted. I feel crestfallen. I will probably begin looking for other jobs soon.
 A: A simple example that maybe your manager could understand: 
Let's say you have Coin Y and Coin X, and you flip each of them 100 times. Then you want to predict whether getting a heads with Coin X (IV) can increase the chance of getting a heads with Coin Y (DV). 
Without sorting, the relationship will be none, because Coin X's outcome shouldn't affect the Coin Y's outcome. With sorting, relationship will be nearly perfect. 
How does it make sense to conclude that you have a good chance of getting a heads on a coin flip if you have just flipped a heads with a different coin?
A: Plenty of good counter examples in here. Let me just add a paragraph about the heart of the problem.
You are looking for a correlation between $X_i$ and $Y_i$. That means that $X$ and $Y$ both tend to be large for the same $i$ and small for the same $i$. So a correlation is a property of $X_1$ linked with $Y_1$, $X_2$ linked with $Y_2$, and so on. By sorting $X$ and $Y$ independently you (in most cases) lose the pairing. $X_1$ will no longer be paired up with $Y_1$. So the correlation of the sorted values will not measure the connection between $X_1$ and $Y_1$ that you are after.
Actually, let me add a paragraph about why it "works" as well.
When you sort both lists, let's call the new sorted list $X_a$, $X_b$, and so on, $X_a$ will be smallest $X$ value, and $Y_a$ will be the smallest Y value. $X_z$ will be the largest $X$ and $Y_z$ will be the largest $Y$. Then you query the new lists if small and large value co occur. That is, you ask if $X_a$ is small when $Y_a$ is small. Is $X_z$ large when $Y_z$ is large? Of course the answer is yes, and of course we will get almost perfect correlation. Does that tell you anything about $X_1$'s relationship with $Y_1$? No.
A: Actually, the test that is described (i.e. sort the X values and the Y values independently and regress one against the other) DOES test something, assuming that the (X,Y) are sampled as independent pairs from a bivariate distribution.  It just isn't a test of what your manager wants to test.  It is essentially checking the linearity of a QQ-plot, comparing the marginal distribution of the Xs with the marginal distribution of the Ys.  In particular, the 'data' will fall close to a straight line if the density of the Xs (f(x)) is related to the density of the Ys (g(y)) this way:
$f(x) = g((y-a)/b)$ for some constants $a$ and $b>0$.  This puts them in a location-scale family.  Unfortunately this is not a method to get predictions...
A: Strange that the most obvious counterexample is still not present among the answers in its simplest form.
Let $Y = -X$.
If you sort the variables separately and fit a regression model on such data, you should obtain something like $\hat Y \approx X$ (because when the variables are sorted, larger values of one must correspond to larger values of the other).
This is a kind-of a "direct inverse" of the pattern you might be willing to find here.
A: Actually, let's make this really obvious and simple.  Suppose I conduct an experiment in which I measure out 1 liter of water in a standardized container, and I look at the amount of water remaining in the container $V_i$ as a function of time $t_i$, the loss of water due to evaporation:
Now suppose I obtain the following measurements $(t_i, V_i)$ in hours and liters, respectively:  $$(0,1.0), (1,0.9), (2,0.8), (3,0.7), (4,0.6), (5,0.5).$$  This is quite obviously perfectly correlated (and hypothetical) data.  But if I were to sort the time and the volume measurements, I would get $$(0,0.5), (1,0.6), (2,0.7), (3,0.8), (4,0.9), (5,1.0).$$  And the conclusion from this sorted data set is that as time increases, the volume of water increases, and moreover, that starting from 1 liter of water, you would get after 5 hours of waiting, more than 1 liter of water.  Isn't that remarkable?  Not only is the conclusion opposite of what the original data said, it also suggests we have discovered new physics!  
A: It's a QQ-plot, isn't it? You'd use it to compare the distribution of x vs. y. If you'd plot sorted outcomes of a relationship like $x \sim x^2$, the plot would be crooked, which indicates that $x$ and $x^2$ for some sampling of $x$s have different distributions.
The linear regression is usually less reasonable (exceptions exist, see other answers); but the geometry of tails and of distribution of errors tells you how far from similar the distributions are.
A: You are right. Your manager would find "good" results! But they are meaningless. What you get when you sort them independently is that the two either increase or decrease similarly and this gives a semblance of a good model. But the two variables have been stripped of their actual relationship and the model is incorrect.
A: It is a real art and takes a real understanding of psychology to be able to convince some people of the error of their ways.  Besides all the excellent examples above, a useful strategy is sometimes to show that a person's belief leads to an inconsistency with herself.  Or try this approach.  Find out something your boss believes strongly about such as how persons perform on task Y has no relation with how much of an attribute X they possess.  Show how your boss's own approach would result in a conclusion of a strong association between X and Y.  Capitalize on political/racial/religious beliefs.
Face invalidity should have been enough.  What a stubborn boss.  Be searching for a better job in the meantime.  Good luck.
A: I have a simple intuition why this is actually a good idea if the function is monotone: 
Imagine you know the inputs $x_1, x_2,\cdots, x_n$ and they are ranked, i.e. $x_i<x_{i+1}$ and assume $f:\Re\mapsto\Re$ is the unknown function we want to estimate. You can define a random model $y_i = f(x_i) + \varepsilon_i$ where $\varepsilon_i$ are independently sampled as follows:
$$
\varepsilon_i = f(x_{i+\delta}) - f(x_i)
$$
where $\delta$ is uniformly sampled from the discrete set $\{-\Delta,-\Delta+1, \cdots \Delta-1, \Delta\}$. Here, $\Delta\in\mathbb{N}$ controls the variance. For example, $\Delta=0$ gives no noise, and $\Delta=n$ give independent input and outputs.
With this model in mind, the proposed "sorting" method of you boss makes perfect sense: If you rank the data, you somehow reduce this type of noise and the estimation of $f$ should becomes better under mild assumptions.
In fact, a more advanced model would assume that $\varepsilon_i$ are dependent, so that we cannot observe 2 times the same output. In such a case, the sorting method could even be optimal. This might have strong connection with random ranking models, such as Mallow's random permutations. 
PS: I find it amazing how an apparently simple question can lead to interesting new ways of re-thinking standards model. Please thank you boss!
A: Say you have these points on a circle of radius 5. You calculate the correlation:
import pandas as pd
s1 = [(-5, 0), (-4, -3), (-4, 3), (-3, -4), (-3, 4), (0, 5), (0, -5), (3, -4), (3, 4), (4, -3), (4, 3), (5, 0)]
df1 = pd.DataFrame(s1, columns=["x", "y"])
print(df1.corr())

   x  y
x  1  0
y  0  1

Then you sort your x- and y-values and do the correlation again:
s2 = [(-5, -5), (-4, -4), (-4, -4), (-3, -3), (-3, -3), (0, 0), (0, 0), (3, 3), (3, 3), (4, 4), (4, 4), (5, 5)]
df2 = pd.DataFrame(s2, columns=["x", "y"])
print(df2.corr())

   x  y
x  1  1
y  1  1

By this manipulation, you change a data set with 0.0 correlation to one with 1.0 correlation. That's a problem.
A: Let me play Devil's Advocate here. I think many answers have made convincing cases that the boss' procedure is fundamentally mistaken. At the same time, I offer a counter-example that illustrates that the boss may have actually seen results improve with this mistaken transformation.
I think that acknowledging that this procedure might've "worked" for the boss  could begin a more-persuasive argument: Sure, it did work, but only under these lucky circumstances that usually won't hold. Then we can show -- as in the excellent accepted answer -- how bad it can be when we're not lucky. Which is most of the time. In isolation, showing the boss how bad it can be might not persuade him because he might have seen a case where it does improve things, and figure that our fancy argument must have a flaw somewhere.
I found this data online, and sure enough, it appears that the regression is improved by the independent sorting of X and Y because: a) the data is highly positively correlated, and b) OLS really doesn't do well with extreme (high-leverage) outliers. The height and weight have a correlation of 0.19 with the outlier included, 0.77 with the outlier excluded, and 0.78 with X and Y independently sorted.
x <- read.csv ("https://vincentarelbundock.github.io/Rdatasets/csv/car/Davis.csv", header=TRUE)

plot (weight ~ height, data=x)

lm1 <- lm (weight ~ height, data=x)

xx <- x
xx$weight <- sort (xx$weight)
xx$height <- sort (xx$height)

plot (weight ~ height, data=xx)

lm2 <- lm (weight ~ height, data=xx)

plot (weight ~ height, data=x)
abline (lm1)
abline (lm2, col="red")


plot (x$height, x$weight)
points (xx$height, xx$weight, col="red")


So it appears to me that the regression model on this dataset is improved by the independent sorting (black versus red line in first graph), and there is a visible relationship (black versus red in the second graph), due to the particular dataset being highly (positively) correlated and having the right kind of outliers that harm the regression more than the shuffling that occurs when you independently sort x and y.
Again, not saying independently sorting does anything sensible in general, nor that it's the correct answer here. Just that the boss might have seen something like this that happened to work under just the right circumstances.
A: Sorting the columns of the following table independently also makes it look "better":
name,country
Alice,DE
Daniel,US
Christian,DE
Bernadette,US

->
name,country
Alice,DE
Bernadette,DE
Christian,US
Daniel,US

Now, all females are from Germany, and all males are from the US.
We can now much more nicely predict the gender by just knowing the country. Isn't that great? /s
A: One more example. Imagine that you have two variables, one connected with eating chocolate and second one connected to overall well-being. You have a sample of two and your data looks like below:
$$
\begin{array}{cc} 
 \text{chocolate}    & \text{no happiness} \\ 
 \text{no chocolate} & \text{happiness} \\ 
\end{array}
$$
What is the relation of chocolate and happiness based on your sample? And now, change order of one of the columns - what is the relation after this operation?
The same problem can be approached differently. Say, that you have a bigger sample, with some number of cases and you measure two continuous variables: chocolate consumption per day (in grams) and happiness (imagine that you have some way to measure it). If you are interested if they are related you can measure correlation or use linear regression model, but sometimes in such cases people simply dichotomize one variable and use it as a grouping factor with $t$-test (this is not the best and not recommended approach, but let me use it as an example). So you divide your sample into two groups: with high chocolate consumption and with low chocolate consumption. Next, you compare average happiness in both groups. Now imagine what would happen if you sorted happiness variable independently of grouping variable: all the cases with high happiness would go go high chocolate consumption group, and all the low happiness cases would end up in low chocolate consumption group -- would such hypothesis test have any sens? This can be easily extrapolated into regression if you imagine that instead of two groups for chocolate consumption you have $N$ such groups, one for each participant (notice that $t$-test is related to regression).
In bivariate regression or correlation we are interested in pairwise relations between each $i$-th value of $X$ and $i$-th value of $Y$, changing order of the observations destroys this relation. If you sort both variables that this always leads them to be more positively correlated with each other since it will always be the case that if one of the variables increases, the other one also increases (because they are sorted!).
Notice that sometimes we actually are interested in changing order of cases, we do so in resampling methods. For example, we can intentionally shuffle observations multiple times so to learn something about null distribution of our data (how would our data look like if there was no pairwise relations), and next we can compare if our real data is anyhow better than the randomly shuffled. What your manager does is exactly the opposite -- he intentionally forces the observations to have artificial structure where there was no structure, what leads to bogus correlations.
A: If you want to convince your boss, you can show what is happening with simulated, random, independent $x,y$ data. With R:
n <- 1000

y<- runif(n)
x <- runif(n)

linearModel <- lm(y ~ x)


x_sorted <- sort(x)
y_sorted <- sort(y)

linearModel_sorted <- lm(y_sorted ~ x_sorted)

par(mfrow = c(2,1))
plot(x,y, main = "Random data")
abline(linearModel,col = "red")


plot(x_sorted,y_sorted, main = "Random, sorted data")
abline(linearModel_sorted,col = "red")

Obviously, the sorted results offer a much nicer regression. However, given the process used to generate the data (two independent samples) there is absolutely no chance that one can be used to predict the other.
A: Your intuition is correct: the independently sorted data have no reliable meaning because the inputs and outputs are being randomly mapped to one another rather than what the observed relationship was. 
There is a (good) chance that the regression on the sorted data will look nice, but it is meaningless in context. 
Intuitive example: Suppose a data set $(X = age, Y = height)$ for some population. The graph of the unadulterated data would probably look rather like a logarithmic or power function: faster growth rates for children that slow for later adolescents and "asymptotically" approach one's maximum height for young adults and older. 
If we sort $x, y$ in ascending order, the graph will probably be nearly linear. Thus, the prediction function is that people grow taller for their entire lives. I wouldn't bet money on that prediction algorithm.  
A: I thought about it, and thought there is some structure here based on order statistics. I checked, and seems manager's mo is not as nuts as it sounds
Order Statistics Correlation Coefficient as a Novel Association Measurement With Applications to Biosignal Analysis
http://www.researchgate.net/profile/Weichao_Xu/publication/3320558_Order_Statistics_Correlation_Coefficient_as_a_Novel_Association_Measurement_With_Applications_to_Biosignal_Analysis/links/0912f507ed6f94a3c6000000.pdf

We propose a novel correlation coefficient based on order statistics
  and rearrangement inequality. The proposed coefficient represents a
  compromise between the Pearson's linear coefficient and the two
  rank-based coefficients, namely Spearman's rho and Kendall's tau.
  Theoretical derivations show that our coefficient possesses the same
  basic properties as the three classical coefficients. Experimental
  studies based on four models and six biosignals show that our
  coefficient performs better than the two rank-based coefficients when
  measuring linear associations; whereas it is well able to detect
  monotone nonlinear associations like the two rank-based coefficients.
  Extensive statistical analyses also suggest that our new coefficient
  has superior anti-noise robustness, small biasedness, high sensitivity
  to changes in association, accurate time-delay detection ability, fast
  computational speed, and robustness under monotone nonlinear
  transformations.

A: If he has preselected the variables to be monotone, it actually is fairly robust. Google "improper linear models" and "Robin Dawes"  or "Howard Wainer." Dawes and Wainer talk about alternate wayes of choosing coefficients. John Cook has a short column (http://www.johndcook.com/blog/2013/03/05/robustness-of-equal-weights/) on it.   
