China Southeastern Mathematical Olympiad

One can check that $2011$ is a prime number. Let $$f(x)=(x-1)(x-2)\cdots (x-2010)-(x^{2010}+2010!).$$ For $n\in \{1,2,\dots ,2010\}$, we have $n^{2010}\equiv 1(\mathrm{mod}2011)$ (by Fermat's little theorem), and $2010!\equiv -1(\mathrm{mod}2011)$ (by Wilson's theorem), so $$f(n)=(n-1)(n-2)\cdots (n-2010)\equiv 0(\mathrm{mod}2011).$$ This means that $f(x)\equiv 0(\mathrm{mod}2011)$ have $2010$ roots, but the degree of $f(x)$ is $2009$. Using Lagrange's theorem, we can see that the coefficients can all be divided by $2011$. $$\sum _{i=1}^{n}P(A_i)\text{ is the coefficient of }{x}^{1911}\text{, hence }2011\mid \sum _{i=1}^{n}P(A_i).$$