7. The sum of the reciprocals of all positive integers $n$ that make $\left[\frac{n^{2}}{5}\right]$ a prime number is . $\qquad$
7. $\frac{37}{60}$.

Notice that when $n=1,2,3$, $\left[\frac{n^{2}}{5}\right]$ is not a prime number; when $n=4$, $\left[\frac{n^{2}}{5}\right]=3$ is a prime number; when $n=5$, $\left[\frac{n^{2}}{5}\right]=5$ is a prime number; when $n=6$, $\left[\frac{n^{2}}{5}\right]=7$ is a prime number. When $n \geqslant 8$, we can set $n=5 k \pm r($ positive integer $k \geqslant 2, r=0,1$ or 2$)$. Then $\frac{n^{2}}{5}=\frac{(5 k \pm r)^{2}}{5}$ $=\frac{1}{5}\left(25 k^{2} \pm 10 r k+r^{2}\right)$ $=k(5 k \pm 2 r)+\frac{1}{5} r^{2}$.

Thus, $\left[\frac{n^{2}}{5}\right]=k(5 k \pm 2 r)$.
Since $k \geqslant 2$, we have
$5 k \pm 2 r>2$
$\Rightarrow\left[\frac{n^{2}}{5}\right]=k(5 k \pm 2 r)$ is not a prime number.
The positive integers $n$ that make $\left[\frac{n^{2}}{5}\right]$ a prime number are only $4, 5, 6$, thus, their reciprocal sum is
$$
\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60} .
$$