(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 $a_{n}=\left[(n+1) x_{n}\right](n=2,3, \cdots)$ (the symbol $[x]$ denotes the greatest integer not exceeding $x$). Then $\frac{1}{1005}\left(a_{2}+a_{3}+a_{4}+\cdots+a_{2011}\right)=$ $\qquad$ .
(9) 2013 Hint: Let $f(x)=n x^{3}+2 x-n$, then it is easy to know that when $n$ is a positive integer, $f(x)$ is an increasing function.
Since when $n \geqslant 2$,
$$
\begin{aligned}
f\left(\frac{n}{n+1}\right) & =n \times\left(\frac{n}{n+1}\right)^{3}+2 \times \frac{n}{n+1}-n \\
& =\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$.
Therefore, when $n \geqslant 2$, the equation $n x^{3}+2 x-n=0$ has a unique real root $x_{n}$, and $x_{n} \in$ $\left(\frac{n}{n+1}, 1\right)$. So
$$
\begin{array}{l}
n<(n+1) x_{n}<n+1, \\
a_{n}=\left[(n+1) x_{n}\right]=n .
\end{array}
$$

Therefore
$$
\begin{aligned}
& \frac{1}{1005}\left(a_{2}+a_{3}+a_{4}+\cdots+a_{2011}\right) \\
= & \frac{1}{1005}(2+3+4+2011) \\
= & \frac{(2+2011) \times 2010}{2 \times 1005} \\
= & 2013 .
\end{aligned}
$$