Team Selection Test

The answer: $n=12$. We numerate the piles according to their weights in increasing order. Let $S_i$ and $S_i^{\prime }$ be the weights of pile number $i$ before and after removing of two stones. Then $S_1<S_2<\cdots <S_n$ and $S_1^{\prime }>S_2^{\prime }>\cdots >S_n^{\prime }$. Let $S_1-S_1^{\prime }=x$. Then $S_2-S_2^{\prime }\ge x+2,\dots ,S_n-S_n^{\prime }\ge x+2(n-1)$. Now note that the heaviest stone removed from pile number one weights at least $S_1/2016$ (it is not lighter than the average weight of all remaining stones). Therefore, $$S_1-S_1^{\prime }=x\ge \frac{S_1}{2016}+1.$$ Similarly, the lightest stone removed from pile number $n$ is at most $S_n^{\prime }/2016$. Therefore, $$\frac{S_n^{\prime }}{2016}+25\ge S_n-S_n^{\prime }\ge x+2(n-1).$$ These two inequalities imply that $$\frac{S_n^{\prime }-S_1^{\prime }}{2016}+24\ge 2(n-1)\text{ and }n<13.$$ Let us show that $n=12$ is possible. Below $m$ denotes the stone weighted $m$. It can be readily verified that if for $k=1,2,\dots ,12$ the pile number $k$ consists of $1$ stone $2k$, $2004+k$ stones $24$ and $13-k$ stones $25$ then the conditions are satisfied. Done.