9. Given the sequence $\left\{a_{n}\right\}$, where $a_{n+1}=\frac{2005 a_{n}}{2003 a_{n}+2005}, a_{1}=1$, then $a_{2006}=$ $\qquad$
9. $\frac{1}{2004}$.

Simplified solution: From $a_{n+1}=\frac{2005 a_{n}}{2003 a_{n}+2005}$, we get $\frac{2005}{a_{n+1}}-\frac{2005}{a_{n}}=2003$. Therefore, $\frac{2005}{a_{n}}=\frac{2005}{a_{1}}+2003(n-$
$1)=2005+2003(n-1)$. Thus, $\frac{2005}{a_{2006}}=2005+2003 \times 2005=2005 \times 2004$, hence $a_{2006}=\frac{1}{2004}$.