9. For a positive integer $n$, let $x_{n}$ be the real root of the equation
$$
n x^{3}+2 x-n=0
$$

with respect to $x$, and let
$$
a_{n}=\left[(n+1) x_{n}\right](n=2,3, \cdots),
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
\frac{1}{1005}\left(a_{2}+a_{3}+\cdots+a_{2011}\right)=
$$
$\qquad$
9.2013.

Let $f(x)=n x^{3}+2 x-n$.
It is easy to see that when $n$ is a positive integer, $f(x)$ is an increasing function.
When $n \geqslant 2$,
$$
\begin{array}{l}
f\left(\frac{n}{n+1}\right)=n\left(\frac{n}{n+1}\right)^{3}+2 \times \frac{n}{n+1}-n \\
=\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$.
Therefore, when $n \geqslant 2$, the equation $n x^{3}+2 x-n=0$ has a unique real root $x_{n}$, and $x_{n} \in\left(\frac{n}{n+1}, 1\right)$.
Thus, $n<(n+1) x_{n}<n+1$,
$$
a_{n}=\left[(n+1) x_{n}\right]=n \text {. }
$$

Then $\frac{1}{1005} \sum_{n=2}^{2011} a_{n}=\frac{1}{1005} \sum_{n=2}^{2011} n=2013$ :