11. (15th "Hope Cup" Senior High School Competition) Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{n+1}=\frac{\sqrt{3} a_{n}-1}{a_{n}+\sqrt{3}}$ $\left(n \in \mathbf{N}^{*}\right)$, then $a_{2004}=$ $\qquad$ .
11. Given $a_{1}=1, a_{n+1}=\frac{\sqrt{3} a_{n}-1}{a_{n}+\sqrt{3}}$, we have
$$
\begin{array}{l}
a_{2}=\frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{-2 \sqrt{3}+4}{2}=2-\sqrt{3}, \\
a_{3}=\frac{\sqrt{3}(2-\sqrt{3})-1}{(2-\sqrt{3})+\sqrt{3}}=\frac{2 \sqrt{3}-4}{2}=-2+\sqrt{3}, \\
a_{1}=\frac{\sqrt{3}(-2+\sqrt{3})-1}{(-2+\sqrt{3})+\sqrt{3}}=\frac{-2 \sqrt{3}+2}{2 \sqrt{3}-2}=-1, \\
a_{5}=\frac{-\sqrt{3}-1}{-1+\sqrt{3}}=-\frac{2 \sqrt{3}+4}{2}=-2-\sqrt{3}, \\
a_{6}=\frac{\sqrt{3}(-2-\sqrt{3})-1}{(-2-\sqrt{3})+\sqrt{3}}=\frac{-2 \sqrt{3}-4}{-2}=2+\sqrt{3}, \\
a_{7}=\frac{\sqrt{3}(2+\sqrt{3})-1}{(2+\sqrt{3})+\sqrt{3}}=\frac{2 \sqrt{3}+2}{2 \sqrt{3}+2}=1 .
\end{array}
$$

Thus, the sequence $\left\{a_{n}\right\}$ is a periodic sequence with period $T=6$,
and $2004 \div 6=334$,
so $a_{2004}=a_{6}=2+\sqrt{3}$.
Another solution: Let $a_{n}=\tan \theta_{n},-\frac{\pi}{2}<\theta_{n}<\frac{\pi}{2}(n=1.2,3, \cdots)$, using trigonometric formulas, we get
$$
\tan \theta_{n+1}=u_{n-1}=\frac{\sqrt{3} a_{n}-1}{a_{n}+\sqrt{3}}=\frac{a_{n}-\frac{1}{\sqrt{3}}}{1+\frac{a_{n}}{\sqrt{3}}}=\frac{\tan \theta_{n}-\tan 30^{\circ}}{1+\tan \theta_{n} \cdot \tan 30^{\circ}}=\tan \left(\theta_{n}-30^{\circ}\right),
$$

and $-\frac{\pi}{2}<\theta_{w}<\frac{\pi}{2}$,
so $\theta_{n+1}=\theta_{n}-30^{\circ}, n=1,2,3, \cdots$
Thus, $\left\{\theta_{\infty}\right\}$ is an arithmetic sequence with a common difference of $-30^{\circ}$, where $\theta_{1}=\arctan 1=45^{\circ}$,
Therefore, $\theta_{n}=45^{\circ}-(n-1) \times 30^{\circ}, n=1,2,3, \cdots$
Hence, $\theta_{2001}=45^{\circ}-2003 \times 30^{\circ}=45^{\circ}-167 \times 360^{\circ}+30^{\circ}=75^{\circ}+167 \times 360^{\circ}$,
$$
a_{2004}=\tan \theta_{2004}=\tan 75^{\circ}=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \cdot \tan 30^{\circ}}=2+\sqrt{3} .
$$