Let $N$ be a positive integer. Two persons play the following game. The first player writes a list of positive integers not greater than 25, not necessarily different, such that their sum is at least 200. The second player wins if he can select some of these numbers so that their sum $S$ satisfies the condition $200-N \leqslant S \leqslant 200+N$. What is the smallest value of $N$ for which the second player has a winning strategy?

Answer: $N=11$.
If $N=11$, then the second player can simply remove numbers from the list, starting with the smallest number, until the sum of the remaining numbers is less than 212. If the last number removed was not 24 or 25, then the sum of the remaining numbers is at least $212-23=189$. If the last number removed was 24 or 25, then only 24s and 25s remain, and there must be exactly 8 of them since their sum must be less than 212 and not less than $212-24=188$. Hence their sum $S$ satisfies $8 \cdot 24=192 \leqslant S \leqslant 8 \cdot 25=200$. In any case the second player wins.

On the other hand, if $N \leqslant 10$, then the first player can write 25 two times and 23 seven times. Then the sum of all numbers is 211, but if at least one number is removed, then the sum of the remaining ones is at most 188 - so the second player cannot win.