Example 7 Let $f(x)$ be a polynomial with integer coefficients, $f(0)=$
11. 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 $\qquad$ (6)
$(2010$, Xin Zhi Cup Shanghai High School Mathematics Competition)
Let $g(x)=f(x)-2010$.
$$
\begin{array}{l}
\text { By } f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010, \text { we have } \\
g\left(x_{1}\right)=g\left(x_{2}\right)=\cdots=g\left(x_{n}\right)=0,
\end{array}
$$

That is, $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ distinct integer roots of the integer-coefficient polynomial $g(x)$. Therefore,
$$
g(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right) q(x) \text {, }
$$
where $q(x)$ is an integer-coefficient polynomial.
Then $g(0)=f(0)-2010$
$$
=11-2010=-1999 \text {. }
$$

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

Therefore, the maximum value of $n$ is 3.