11. Given the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{0}=0, a_{1}=1$, for all positive integers $n$, there is $a_{n+1}=2 a_{n}+$ $2007 a_{n-1}$, find the smallest positive integer $n$ such that $2008 \mid a_{n}$.
11. Solution: Let $m=2008, a_{n+1}=2 a_{n}+2007 a_{n-1}$, the characteristic equation is $\lambda^{2}-2 \lambda-2007=0$, with characteristic roots $1 \pm \sqrt{m}$. Combining $a_{0}=0, a_{1}=1$, we get $a_{n}=\frac{1}{2 \sqrt{m}}\left((1+\sqrt{m})^{n}-(1-\sqrt{m})^{n}\right)$. By the binomial theorem, $a_{n}=\frac{1}{2 \sqrt{m}}\left(\sum_{k=0}^{n} C_{n}^{k} m^{\frac{k}{2}}-\sum_{k=0}^{n}(-1)^{k} C_{n}^{k} m^{\frac{k}{2}}\right)$.
When $n$ is odd, $a_{n}=C_{n}^{1}+C_{n}^{3} m+\cdots+C_{n}^{n-2} m^{\frac{n-3}{2}}+C_{n}^{n} m^{\frac{n-1}{2}}$;
When $n$ is even, $a_{n}=C_{n}^{1}+C_{n}^{3} m+\cdots+C_{n}^{-3} m^{\frac{n-4}{2}}+C_{n}^{n-1} m^{\frac{n-2}{2}}$.
Thus, $m\left|a_{n} \Leftrightarrow m\right| C_{n}^{1}$, i.e., $2008 \mid n$, so the smallest positive integer satisfying the condition is 2008.