7. Let $f(x)$ be a polynomial with integer coefficients, $f(0)=11$, and there exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$
f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010 .
$$

Then the maximum value of $n$ is
7.3.

Let $g(x)=f(x)-2010$. Then $x_{1}, x_{2}, \cdots, x_{n}$ are all roots of $g(x)=0$.
Thus, $g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right) \cdot q(x)$,

where $q(x)$ is a polynomial with integer coefficients. Therefore,
$$
\begin{array}{l}
g(0)=11-2010=-1999 \\
=\prod_{i=1}^{n}\left(-x_{i}\right) q(0) .
\end{array}
$$

Since 1999 is a prime number, -1999 can be at most the product of three different integers, i.e., $-1 \times 1 \times 1999$, hence $n \leqslant 3$.
When $n=3$,
$$
\begin{array}{l}
g(x)=(x+1)(x-1)(x+1999), \\
f(x)=(x+1)(x-1)(x+1999)+2010 .
\end{array}
$$