China Southeastern Mathematical Olympiad

For each 99-element subset, $A_i=\{a_1,a_2,\dots ,a_{99}\}$ of $\{1,2,\dots ,2010\}$ uniquely corresponds to a 99-element subset $B_i=\{b_1,b_2,\dots ,b_{99}\}$ of $\{1,2,\dots ,2010\}$ by $b_k=2011-a_k$, $k=1,2,\dots ,99$. Since ${\sum }_{k=1}^{99}(a_k+b_k)=99\times 2011$ is odd, we see that $A_i,B_i$ are different subsets of $\{1,2,\dots ,2010\}$. When $A_i$ take all 99-element subsets of $\{1,2,\dots ,2010\}$, so do $B_i$. Moreover $$\begin{array}{rl}P(A_i)+P(B_i)& =a_1{a}_2\cdots a_{99}+(2011-a_1)(2011-a_2)\cdots (2011-a_{99})\\ & \equiv a_1{a}_2\cdots a_{99}+(-a_1)(-a_2)\cdots (-a_{99})(\mathrm{mod}2011)\\ & \equiv 0(\mathrm{mod}2011).\end{array}$$ Thus, $$2\sum _{i=1}^{n}P(A_i)=\sum _{i=1}^{n}P(A_i)+\sum _{i=1}^{n}P(B_i)\equiv 0(\mathrm{mod}2011),$$ hence $2011\mid {\sum }_{i=1}^{n}P(A_i)$.

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Alternative solution.

Let $f(n)=(n-1)(n-2)\cdots (n-2010)-n^{2010}-2010!$, where $n\in \mathbb{Z}$. Since 2011 is prime, by Fermat's Little Theorem, $n^{2010}\equiv 1(\mathrm{mod}2011)$. By Wilson's Theorem, we have $2010!\equiv -1(\mathrm{mod}2011)$. Thus, (i) If $2011\nmid n$, then $$f(n)\equiv (n-1)(n-2)\cdots (n-2010)\equiv 0(\mathrm{mod}2011).$$ (ii) If $2011\mid n$, then $$\begin{array}{rl}f(n)& \equiv (2011-1)(2011-2)\cdots (2011-2010)-2011^{2010}-2010!\\ & \equiv 2010!-2010!(\mathrm{mod}2011)\\ & \equiv 0(\mathrm{mod}2011).\end{array}$$ So $f(n)\equiv 0(\mathrm{mod}2011)$ has 2011 solutions in the sense of $\mathrm{mod}2011$. Since $f(n)$ is a polynomial of order 2009, and for all $n\in \mathbb{Z}$, $2011\mid f(n)$, we see that each coefficient of $f(n)$ can be divided by 2011. Turn to the original problem, ${\sum }_{i=1}^{n}P(A_i)$ is the coefficient of term with order 1911 of $f(n)$, thus $2011\mid {\sum }_{i=1}^{n}P(A_i)$.