1. If 2009 numbers $a_{1}, a_{2}, \cdots, a_{2009}$ satisfy: $a_{1}$ $=3, a_{n}^{2}-\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right) a_{n}+\frac{1}{2009}=0$ (where $n=$ $2,3,4, \cdots, 2009$ ), then the maximum value that $a_{2009}$ can reach is
1. $\left(a_{2009}\right)_{\max }=\frac{1}{3} \cdot 2009^{2007}$.

Solution: $a_{n}=$
$$
\begin{array}{l}
\frac{\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right) \pm \sqrt{\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right)^{2}-\frac{4}{2009}}}{2} \\
=\frac{\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right) \pm\left|\frac{a_{n-1}}{2009}-\frac{1}{a_{n-1}}\right|}{2},
\end{array}
$$

Therefore, $a_{n}=\frac{a_{n-1}}{2009}\left(a_{n-1}>\sqrt{2009}\right)$ or $a_{n}=\frac{1}{a_{n-1}}\left(a_{n-1}\right.$ $<\sqrt{2009})$,

Hence $\left(a_{2008}\right)_{\min }=3\left(\frac{1}{2009}\right)^{2007}, \therefore\left(a_{2009}\right)_{\max }=\frac{1}{3} \cdot 2009^{2007}$.