, 2010

Assume that $m$ divides $f(n)$ for some integer $n$ and some $2 \le m \le 2010$. As $f(1) = 2011$ and $2011$ is a prime number, $m$ cannot be a divisor of $f(1)$, so we may restrict ourselves to the case $n \ne 1$.
In this case, we can write $f(n)$ as
$$f(n) = \frac{n^{2011} - 1}{n - 1}.$$
Let $p$ be a prime divisor of $m$. Then we have $p \mid f(n) \mid n^{2011} - 1$, which results in
$$n^{2011} \equiv 1 \pmod{p}. \qquad (1)$$
This immediately implies that $n$ and $p$ are coprime.
By (1), the order $\text{ord}_p(n)$ of $n$ modulo $p$, i.e., the smallest positive exponent $k$ such that $n^k \equiv 1 \pmod{p}$, is a divisor of $2011$. As $2011$ is prime, this results in $\text{ord}_p(n) \in \{1, 2011\}$.
If $\text{ord}_p(n) = 1$, we have $n \equiv 1 \pmod{p}$, which results in $0 \equiv f(n) \equiv 2011 \pmod{p}$ (by the original definition of $f(n)$). This implies $p \mid 2011$ and therefore $p = 2011$, contradiction.
We conclude that $\text{ord}_p(n) = 2011$. As $\text{ord}_p(n)$ always divides $\varphi(p) = p-1$ by Fermat's theorem, we conclude that $p-1$ is a multiple of $2011$. This is a contradiction to $1 < p < 2011$. qed