$14 \cdot 30$ What is the sum of the reciprocals of all natural numbers $n$ that make $\left[\frac{n^{2}}{5}\right]$ a prime number?
(China Beijing High School Grade 1 Mathematics Competition, 1991)
［Solution］Let $m=\left[\frac{n^{2}}{5}\right]$.
(1) When $n=5 k$ (where $k$ is a natural number),
$$
m=\left[\frac{25 k^{2}}{5}\right]=5 k^{2} \text {. }
$$

When $k=1$, $m$ is a prime number, at this time $n=5$.
(2) When $n=5 k+1$ (where $k$ is a non-negative integer),
$$
m=\left[\frac{(5 k+1)^{2}}{5}\right]=5 k^{2}+2 k+\left[\frac{1}{5}\right]=k(5 k+2) \text {. }
$$

When $k=1$, $m$ is a prime number, at this time $n=6$.
(3) When $n=5 k+2$ (where $k$ is a non-negative integer),
$$
m=\left[\frac{(5 k+2)^{2}}{5}\right]=5 k^{2}+4 k=k(5 k+4) \text {. }
$$

When $k=1$, $m=9$ is a composite number, so for all positive integers $k$, $m$ is a composite number.
(4) When $n=5 k+3$ (where $k$ is a non-negative integer),
$$
m=\left[\frac{(5 k+3)^{2}}{5}\right]=(5 k+1)(k+1) \text {. }
$$

When $k=0$, $m=1$, when $k$ is a positive integer, $m$ is a composite number.
(5) When $n=5 k+4$ (where $k$ is a non-negative integer),
$$
m=\left[\frac{(5 k+4)^{2}}{5}\right]=(5 k+3)(k+1) \text {. }
$$

When $k=0$, $m=3$ is a prime number, when $k$ is a positive integer, $m$ is a composite number. At this time $n=$
4.
Therefore, when $n=4,5,6$, $\left[\frac{n^{2}}{5}\right]$ is a prime number.
The sum of the reciprocals of such $n$ is
$$
\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60} \text {. }
$$