8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
a_{n+2}=\frac{2(n+1)}{n+2} a_{n+1}-\frac{n}{n+2} a_{n}(n=1,2, \cdots) .
\end{array}
$$

If $a_{m}>2+\frac{2011}{2012}$, then the smallest positive integer $m$ is . $\qquad$
8.4025.
$$
\begin{array}{l}
\text { Given } a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
\begin{array}{l} 
\Rightarrow a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
= \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdots \frac{1}{3}\left(a_{2}-a_{1}\right) \\
= \frac{1 \times 2}{n(n-1)}=2\left(\frac{1}{n-1}-\frac{1}{n}\right) \\
\Rightarrow a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right) \\
=1+2\left(1-\frac{1}{n}\right)=3-\frac{2}{n} .
\end{array}
\end{array}
$$

Thus, $a_{m}>2+\frac{2011}{2012}$
$$
\begin{array}{l}
\Leftrightarrow 3-\frac{2}{m}>3-\frac{1}{2012} \\
\Leftrightarrow m>4024 .
\end{array}
$$

In summary, the smallest value of $m$ that satisfies the condition is 4025.