8. Let $a_{1}=1, a_{2}=2$, for $n \geqslant 2$ we have
$$
a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} .
$$

If for all positive integers $n \geqslant m$, we have $a_{n}>2+$ $\frac{2008}{2009}$, then the smallest positive integer $m$ is $\qquad$ .
8.4019 .

According to the problem, the recursive relationship becomes
$$
\begin{array}{l}
a_{n+1}-a_{n}=\frac{n-1}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
=\frac{n-1}{n+1}\left(a_{n}-a_{n-1}\right)(a \geqslant 2) .
\end{array}
$$

Therefore, when $n \geqslant 3$,
$$
\begin{array}{l}
a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdot \frac{2}{4} \cdot \frac{1}{3}\left(a_{2}-a_{1}\right) \\
=\frac{2}{n(n-1)}=\frac{2}{n-1}-\frac{2}{n} .
\end{array}
$$

Thus, $a_{n}-a_{2}=1-\frac{2}{n}=\frac{n-2}{n}$.
Hence, $a_{n}=a_{2}+\frac{n-2}{n}=\frac{3 n-2}{n}(n \geqslant 3)$. Let $\frac{3 n-2}{n}>2+\frac{2008}{2009}$, solving gives $n>4018$.
Therefore, the minimum value of $m$ is 4019.