8. Let the integer pair $(m, n)$ satisfy $\frac{m^{2}+m n+n^{2}}{m+2 n}=\frac{13}{3}$. Then $m+2 n=$ $\qquad$ .
8.9.

From the problem, we know that $m+2 n \neq 0$.
Thus, the given equation can be written as
$$
3\left(m^{2}+m n+n^{2}\right)=13(m+2 n) \text {. }
$$

Multiplying both sides of the above equation by 4, we get
$$
\begin{array}{l}
12\left(m^{2}+m n+n^{2}\right)=52(m+2 n) \\
\Rightarrow 9 m^{2}+3(m+2 n)^{2}=52(m+2 n) .
\end{array}
$$

Since $(3,52)=1$, it follows that $m+2 n$ must be a multiple of 3. In this case, both terms on the left side of equation (1) are multiples of 9. Therefore, $m+2 n$ is also a multiple of 9.
Furthermore, the first term on the left side of equation (1) is non-negative, hence,
$$
\begin{array}{l}
52(m+2 n) \geqslant 3(m+2 n)^{2} \\
\Rightarrow 0<m+2 n \leqslant \frac{52}{3} \Rightarrow m+2 n=9 .
\end{array}
$$

Substituting into equation (1), we get $(m, n)=(5,2)$ or $(-5,7)$.
Both solutions satisfy the conditions of the problem.