How many of the biggest terms in $\sum_{i=1}^N |X_i|$ add up to half the total? Consider $\sum_{i=1}^N |X_i|$
where $X_1, \ldots, X_N$ are i.i.d. and the CLT holds.
How many of the biggest terms add up to half the total sum ?
For example, 10 + 9 + 8 $\approx$ (10 + 9 + 8 $\dots$ + 1) / 2: 
30 % of the  terms reach about half the total.
Define
$ \qquad\text{sumbiggest( j}; X_1 \dots X_N ) \equiv
    \text{sum of the j biggest of } |X_1| \dots |X_N| $
$ \qquad\text{halfsum}( N ) \equiv
   \text{the smallest j such that sumbiggest( j )} \approx \text{sumbiggest}( N ) / 2 . $
Is there a general asymptotic result for halfsum( $N, \mu, \sigma$ ) ?
A simple, intuitive derivation would be nice.
(A little Monte Carlo suggests that sometimes
halfsum( $N$ ) $\approx N$/4 or so;
that is, the biggest 1/4 of the $X_i$ add up to 1/2 the total.
I get 0.24 $N$ for halfnormal,
0.19 $N$ for exponential, for $N$ = 20, 50, 100.)
 A: No, there isn't a general asymptotic result.  Let $x_{[1]} \dots x_{[N]}$ be the ordered $x_i$, where $x_{[1]}$ is the largest.
Consider the following two examples:
1) $P(x=0) = 1$.  Clearly the CLT holds.  You only need $M=1$ observation for $\sum_{j=1}^M|x_{[j]}| \ge \frac{1}{2} \sum_N|x_i|$.  
2) $P(x=1) = 1$.  Clearly the CLT holds.  You need $M=\lceil N/2\rceil$ observations for $\sum_{j=1}^M|x_{[j]}| \ge \frac{1}{2} \sum_N|x_i|$.
For a nontrivial example, the Bernoulli distribution:
3) $P(x=1) = p,\space P(x=0) = 1-p$.  Once again the CLT holds.  You need $\lceil pN/2\rceil $ of the observations to meet your conditions.  By varying $p$ between 0 and 1, you can get as close to example 1 or example 2 as you like.
A: Here is a crude argument giving a slightly different estimate for uniformly distributed random variables.  Suppose the $X_i$ are continuous random
variables uniformly distributed on $[0,1]$.  Then, $\sum_i X_i$ has mean value $N/2$.  Assume that by a surprising and totally unbelievable coincidence, 
the sum is exactly equal to $N/2$.  So we want to estimate how 
many of the largest values of $X$ sum up to $N/4$ or more.  Now, the histogram 
of $N$ samples ($N$ very large) drawn from the uniformm distribution $U[0,1]$
is roughly flat from $0$ to $1$, and so for any $x$, $0 < x < 1$, there are 
$(1-x)N$ samples distributed roughly uniformly between $x$ to $1$.  These
samples have average value $(1+x)/2$ and sum equal to 
$(1-x)N(1+x)/2) = (1-x^2)N/2$.  The sum exceeds $N/4$ for $x \leq 1/\sqrt{2}$.
So, the sum of $(1-1/\sqrt{2})N \approx 0.3N$ largest samples exceeds
$N/4$.
You could try and generalize this a bit. If $\sum_i X_i = Y$, then for
any given $Y$, we want $x$ to be such that $(1-x^2)N/2 = Y/2$ where
$Y$ is normal with mean $N/2$ and variance $N/12$.  Thus, conditioned
on a value of $Y$, $x = \sqrt{1-(Y/N)}$. Multiply by the
density of $Y$ and integrate (from $Y=0$ to $Y=N$) to find the 
average number of largest samples that will exceed half the random sum.
A: Let's assume X has just positive values to get rid of the absolute value.
Without an exact prove, I think you have to solve for k
$(1-F_{X}(k))E(X|X>=k)= \frac{1}{2} E(X)$ with F being the cumulative distribution function for X
and then the answer is given by taking the $n(1-F_X(k))$ highest values.
My logic is that asymtopically the sum of all values higher than k should be about
$n(1-F_{X}(k))E(X|X>=k)$
and asymtopically half the total sum is about
$\frac{1}{2}nE(X)$. 
Numerical simulation show that the result holds for the uniform case (uniform in $[0,1]$) where $F(k)=k$ and I get $k=\sqrt(\frac{1}{2})$. I am not certain if the result always hold or if it can be simplified further, but I think it really depends on the distribution function F.
