Without loss of generality, assume $f(0)$ is an integer (otherwise, consider instead the function $f(x)-f(0)$). We need only prove that $f(2011)$ and $f(100)$ are both integers.

We use mathematical induction to prove a more general proposition: if an $n$-th degree polynomial $f(x)$ satisfies that $f(0), f(1), \dots, f(n)$ are all integers, then for any integer $x$, $f(x)$ is also an integer.

First, for the base case $n=0$, $f(x)$ is a constant polynomial, so the proposition clearly holds.

Suppose the proposition holds for all $n < k$. Then for a $k$-th degree polynomial $f(x)$ such that the values $f(0), f(1), \dots, f(k)$ are all integers, consider the function $g(x) = f(x+1) - f(x)$. It is easy to see that $g(x)$ is a polynomial of degree $< k$, and that $g(0), g(1), \dots, g(k-1)$ are all integers. By the induction hypothesis, $g(x)$ takes integer values whenever $x$ is an integer. Since for nonnegative integers $x$ we have $f(x) = f(0) + \sum_{i=0}^{x-1} g(i)$, and for negative integers $x$ we have $f(x) = f(0) - \sum_{i=x}^{-1} g(i)$. Therefore, when $x$ is an integer, $f(x)$ is also an integer.