Five. (15 points) Let $\left\{a_{n}\right\}$ be an integer sequence, and it satisfies: for any $n\left(n \in \mathbf{N}_{+}\right)$, we have
$$
(n-1) a_{n+1}=(n+1) a_{n}-2(n-1) \text {, }
$$

and $2008 \mid a_{2000}$. Find the smallest positive integer $n(n \geqslant 2)$, such that $2008 \mid a_{n}$.
When $n=1$, we have $a_{1}=0$.
When $n \geqslant 2$, the original equation transforms to
$$
a_{n+1}=\frac{n+1}{n-1} a_{n}-2 \text{. }
$$

Let $b_{n}=\frac{a_{n}}{n-1}$, then $n b_{n+1}=(n+1) b_{n}-2$.
Thus, for $n \geqslant 2$, we have
$$
b_{n+1}-2=\frac{n+1}{n}\left(b_{n}-2\right) \text{. }
$$

From equation (2), we know
$$
\begin{array}{l}
b_{n}-2=\frac{n}{n-1} \cdot \frac{n-1}{n-2} \cdots \cdot \frac{3}{2}\left(b_{2}-2\right)=\left(\frac{b_{2}}{2}-1\right) n \\
\Rightarrow a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] .
\end{array}
$$

Given $2008 \mid a_{2000}$, we have $\frac{a_{2}}{2} \in \mathbf{Z}$.
Thus, let $a_{2}=2 k(k \in \mathbf{Z})$.
From equation (3), we get $a_{n}=(n-1)[(k-1) n+2]$.
Given $2008 \mid a_{2007}$
$$
\begin{array}{l}
\Rightarrow 2006(2007 k-2005) \equiv 0(\bmod 2008) \\
\Rightarrow k \equiv 3(\bmod 1004) \\
\Rightarrow a_{n}=(n-1)[(1004 m+2) n+2](m \in \mathbf{Z}) .
\end{array}
$$

Then $20081 a_{n}$
$$
\begin{array}{l}
\Leftrightarrow 20081(n-1)[(1004 m+2) n+2] \\
\Rightarrow 1004 \mid(n-1)(n+1) .
\end{array}
$$

Thus, $2 \mid n$.
Let $n=2 l+1(l \in \mathbf{Z}) \Rightarrow 251 \mid l(l+1)$.
Since 251 is a prime number, and $(l, l+1)=1$, we have
$$
l+1 \geqslant 251 \Rightarrow l_{\text {min }}=250 \Rightarrow n_{\text {min }}=501 .
$$