60th Belarusian Mathematical Olympiad

If Nick fixes the number $n$ and defines himself as the first player, then Mary wins if she sets $m=11-n$. Indeed, if Nick removes $n$ ($m$) stones, then Mary removes $m$ ($n$) stones. So exactly $11=m+n$ stones are removed from the heap after each pair of moves (Nick - Mary). Since $25201=2291\cdot 11$, Mary wins.

If Nick fixes the number $n$ and defines Mary as the first player, then Mary wins if she sets $m=1$ for $n=2,3,4,5,6,7,8,9$ and her first move is to remove 1 stone from the heap. $25200$ stones remain in the heap. It is easy to see that $2520$ is divide by $3,4,5,6,7,8,9,10$, i.e. $25200$ is divided by $n+m$ for the fixed value of $n$ and $m$. Therefore, Mary wins if she uses the symmetric strategy: when Nick removes $n$ ($m$) stones Mary removes $m$ ($n$) stones. So exactly $m+n$ stones are removed from the heap after each pair of moves (Nick - Mary).

If Nick sets $n=1$, then Mary can choose any number from $2,3,4,5,6,7,8,9$ for $m$, and play in the same way as above (first move is $1$ stone, and symmetric strategy).

If Nick sets $n=10$, then Mary wins if she sets $m=7$ and her first move is to remove $7$ stones from the heap. $25194$ stones remain in the heap. It is easy to see that $25194=1482\cdot 17$ is divided by $17$, i.e. $25194$ is divided by $n+m$. To win Mary can use the symmetric strategy.