9.5. If there are several piles of stones on the table, it is considered that there are many stones on the table if one can find 50 piles and number them from 1 to 50 such that the first pile has at least one stone, the second pile has at least two stones, ..., the fiftieth pile has at least fifty stones. Suppose initially there are 100 piles, each with 100 stones. Find the largest \( n \leqslant 10000 \) such that after removing any \( n \) stones from the initial piles, there will still be many stones on the table. (When removing stones, a pile does not split into several.) (D. Khramov)
Answer. $n=5099$.

Solution. If 51 piles are completely removed, obviously, there will not be many stones left. Therefore, the desired value of $n$ is less than 5100. (Alternatively, one can remove 51 stones from each pile.)

It remains to show that after removing any $n=5099$ stones, there will still be many stones left. Let $a_{1}, a_{2}, \ldots$, $a_{100}$ be the number of stones left in the piles, respectively; we can assume that $0 \leqslant a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{100} \leqslant 100$. We need to show that $a_{i+50} \geqslant i$ for $i=1,2, \ldots, 50$, meaning that the piles from 51 to 100 meet the requirements.

Suppose this is not the case, i.e., $a_{i+50} \leqslant i-1$ for some $i \leqslant 50$. This means that each of the first $i+50$ piles contains no more than $i-1$ stones, so at least $101-i$ stones have been removed from each of them. Therefore, the total number of removed stones is at least $(i+50)(101-i)=5100-(i-1)(i-50) \geqslant 5100$. This is a contradiction.

Comment. An example of how removing 5100 stones can result in no many stones left - 1 point.

Proof that removing any 5099 stones will still leave many stones - 6 points.