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.