11. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

Try to find $\left[a_{2009}^{2}\right]$.
11. From $a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}$, we get $a_{n+1}^{2}=\frac{a_{n}^{2}}{n^{2}}+\frac{n^{2}}{a_{n}^{2}}+2$. Given $a_{1}=1$, we have $a_{2}^{2}=4, a_{3}^{2}=4, a_{4}^{2}=4+\frac{25}{36}$.
Next, we use mathematical induction to prove:
When $n \geqslant 4$, $n+\frac{2}{n}\frac{n+1}{n^{2}}+\frac{n^{2}}{n+1}+2 \\
=\frac{n+1}{n^{2}}+\frac{1}{n+1}+(n-1)+2 \\
>n+1+\frac{2}{n+1} .
\end{array}
$

On the other hand,
$
\begin{array}{l}
a_{n+1}^{2}=\frac{a_{n}^{2}}{n^{2}}+\frac{n^{2}}{a_{n}^{2}}+2\left(n^{2}+2\right)^{2} \\
\Leftrightarrow n^{4}>4 n^{2}+4 . \\
\text { When } n \geqslant 4 \text {, } n^{4} \geqslant 4 \times 4 n^{2}>4 n^{2}+4, \text { (3) (3) }
\end{array}
$

is true.
From equations (2) and (3), we get $a_{n+1}^{2}<n+2$.
In summary, $n+1+\frac{2}{n+1}<a_{n+1}^{2}<n+2$.
By the principle of induction, when $n \geqslant 4$, equation (1) always holds. Therefore, $\left[a_{2009}^{2}\right]=2009$.