if there's a box with 2 balls, surely there won't be any disjoint class of boxes with 98 balls, but you can't say now that there's no class with 96, since if you add the box with 2 balls, it is not disjoint from itself
Let's prove a more generalized version.
Given $2n$ balls distributed in $n$ boxes, with $n>1$ and even, such that every box is not empty and has at most $n$ balls in it, then there exists a subset of the box whit exactly $n$ balls.
First of all, if all the boxes contains exactly 2 balls, it's easy, so let's assume there's a box with more than 2 balls, and consequentially one with 1 ball.
Order the boxes in decreasing order and start picking them from the first. Continue until the balls count get over $n$.
Let's say you've picked the first $k$ boxes, $a$ is the number of balls in them, and $b$ is the number of balls in the $k+1$-th box, with $\;a<n\;$ but $\;a+b>n\;$ (if it's equal to $n$, you've finished).
Surely $b>1$, and if $b=2$, then $a=n-1$, and we can add a box with a ball.
If $b>2$, it's easy to show that there are at least $b-2$ boxes with exactly 1 ball, so if $a+b-2\ge n$, we've finished, by picking the right number of boxes with 1 ball.
The only case left is $a+b-2=n-1$, that is $a+b=n+1$.
If $a\ne 0$, then $k>0$, and so there's surely at least another box with 1 ball (different from the $b-2$ ones), and we're done.
If $a=0$, then $b=n+1$, impossible.
The claim is that at least one of the $k$ boxes contain at least $\lceil N/k\rceil$ objects.
The proof goes by contradiction: Suppose the claim is false, then each box must have strictly less than $\lceil N/k\rceil$ objects, i.e., at most $\lceil N/k\rceil-1$ objects (the greatest integer strictly less than $n$ is $n-1$)
Now, then since there are $k$ boxes and each box has objects $\leq\lceil N/k\rceil-1$, the total number of objects from all the boxes is $\leq k(\lceil N/k\rceil-1)\lt k\cdot N/k=N$ (we use $\lceil x\rceil-1\lt x$) which gives us a contradiction since the total number of objects from all the boxes cannot be strictly less than $N$ (since $N$ is the total number of objects from all the boxes).
Notice the one single $\lt$ in the chain of inequalities in the penultimate step which makes the overall inequality strict, i.e, gives us $N\lt N$
Here's an informal (intuitive) way to interpret the theorem:
We usually look at the worst case scenario. If we want to keep the number of objects in each box as less as possible when putting the objects in the boxes, we can do this by putting one object in each box and then going over to the next box and not fill a previous box unless all the boxes have the same number of objects. We put an object in the 1st box, then one in the 2nd,.. and so on till the $k$th box until all the boxes have one object and then we repeat the above.
If $N$ is a multiple of $k$, then we'd have $N/k$ objects in each box at the end.
Otherwise, we'll have the first $x$ boxes with $\lceil N/k\rceil$ objects where $x$ is the remainder when $N$ is divided by $k$ and the rest of the boxes will have exactly one less object than the first $x$ boxes.
Best Answer
Let's say the max number of hairs is 500,000 (which i think is rreeaallyy high..) so the claim is true for any city with 500,000+ (non bald) people. (there should be an assumption that there are more then 500,000 non bald people). proof: try to give each person a different number of hairs, 1,2,3... After 500,000 people, you've run out of different numbers. meaning the 500,001th person will have to get a number of hairs which is equal to some other person's number of hairs. These two are in the same pigeonhole