Example 13 Find all values of $a$ such that the roots of the equation $a^{2} x^{2}+a x+1-7 a^{2}=0$ are integers.
(1992, Shanghai Junior High School Mathematics Competition)
Solution: Transform the original equation into a quadratic equation in $a$:
$$
\begin{array}{l}
\left(x^{2}-7\right) a^{2}+x a+1=0 . \\
\because \Delta=x^{2}-4\left(x^{2}-7\right)=28-3 x^{2} \geqslant 0, \\
\therefore x^{2} \leqslant \frac{28}{3} .
\end{array}
$$

Since $x$ is an integer,
$$
\therefore x^{2}=0,1,4,9 \text {. }
$$

If $a=\frac{-x \pm \sqrt{28-3 x^{2}}}{2\left(x^{2}-7\right)}$,
$\therefore$ when $x=1$, $a=\frac{1}{2}$ or $-\frac{1}{3}$;
when $x=-1$, $a=\frac{1}{3}$ or $-\frac{1}{2}$;
when $x=2$, $a=1$ or $-\frac{1}{3}$;
when $x=-2$, $a=\frac{1}{3}$ or $-1$;
when $x=3$, $a=-\frac{1}{2}$ or $-1$;
when $x=-3$, $a=\frac{1}{2}$ or $1$;
when $x=0$, $a=\frac{\sqrt{7}}{7}$ or $-\frac{\sqrt{7}}{7}$.
When $a>0$ and $a=1$, $x=2$ or -3; when $a=\frac{1}{2}$, $x=1$ or -3;
when $a=\frac{1}{3}$, $x=-2$ or -1;

Therefore, the sum of the values of $a$ is $1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$.