$9 \cdot 37$ Find the largest real number $\alpha$ such that for any positive integers $m$ and $n$ satisfying $\frac{m}{n}<\sqrt{7}$, we have
$$
\frac{\alpha}{n^{2}} \leqslant 7-\frac{m^{2}}{n^{2}} \text {. }
$$
[Solution] Since
$$
\begin{array}{l}
\frac{m}{n}0, \\
\frac{\alpha}{n^{2}} \leqslant 7-\frac{m^{2}}{n^{2}} \Leftrightarrow \alpha \leqslant 7 n^{2}-m^{2},
\end{array}
$$

the maximum real number sought is
$$
\alpha=\min \left\{7 n^{2}-m^{2} ; n, m \in N \text {, and } 7 n^{2}-m^{2}>0\right\} \text {. }
$$

Since $7 n^{2}$ is divisible by 7, the remainder of $m^{2}$ when divided by 7 can only be $0,1,2,4$, so if $7 n^{2}-m^{2}>0$, then
$$
7 n^{2}-m^{2} \geqslant 3 \text {, }
$$

thus $\alpha \geqslant 3$. Furthermore, when $n=1, m=2$, $7 n^{2}-m^{2}=3$, hence $\alpha=3$.