11. Let $\left\{a_{n}\right\}$ be an integer sequence, and it satisfies: for any $n\left(n \in \mathbf{N}^{*}\right)$, we have $(n-1) a_{n+1}=$ $(n+1) a_{n}-2(n-1)$, and $2008 \mid a_{2007}$, find the smallest positive integer $n(n \geqslant 2)$, such that $2008 \mid a_{n}$.
11. Solution: When $n=1$, we have $a_{1}=0$; when $n \geqslant 2$, the original equation transforms to $a_{n+1}=\frac{n+1}{n-1} a_{n}-2$ (1); let $b_{n}=\frac{a_{n}}{n-1}$, then $n b_{n+1}=(n+1) b_{n}-2$,
thus, when $n \geqslant 2$, we have $b_{n+1}-2=\frac{n+1}{n}\left(b_{n}-2\right)$ (2);
From equation (2), we know
$b_{n}-2=\frac{n}{n-1} \cdot \frac{n-1}{n-2} \cdot \cdots \cdot \frac{3}{2}\left(b_{2}-2\right)=\left(\frac{b_{2}}{2}-1\right) n \Rightarrow a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right]$
Given $2008 \mid a_{2007}$, we have $\frac{a_{2}}{2} \in \mathbf{Z}$, so let $a_{2}=2 k(k \in \mathbf{Z})$.
From equation (3) $\Rightarrow a_{n}=(n-1)[(k-1) n+2]$,
Given $2008 \mid a_{2007} \Rightarrow 2006(2007 k-2005) \equiv 0(\bmod 2008) \Rightarrow k \equiv 3(\bmod 1004)$
$\Rightarrow a_{n}=(n-1)[(1004 m+2) n+2](m \in \mathbf{Z})$,
then $2008\left|a_{n} \Leftrightarrow 2008\right|(n-1)[(1004 m+2) n+2] \Rightarrow 1004 \mid(n-1)(n+1)$, hence 2 does not divide $n$.
Let $n=2 l+1(l \in \mathbf{Z}) \Rightarrow 251 \mid l(l+1)$, since 251 is a prime number, and $(l, l+1)=1$,
so $l+1 \geqslant 251 \Rightarrow l_{\min }=250 \Rightarrow n_{\min }=501$.