14.4.9 * Find the sum of the reciprocals of all natural numbers $n$ that make $\left[\frac{n^{2}}{5}\right]$ a prime number.
When $n=5 k(k=1,2, \cdots)$, $\left[\frac{n^{2}}{5}\right]=5 k^{2}$, and only when $k=1$, i.e., $n=5$, $\left[\frac{n^{2}}{5}\right]$ is a prime number. When $n=5 k+1(k=0,1,2, \cdots)$, $\left[\frac{n^{2}}{5}\right]=5 k^{2}+2 k=k(5 k+2)$, only when $k=1$, i.e., $n=6$, $\left[\frac{n^{2}}{5}\right]$ is a prime number. When $n=5 k+2(k=0,1,2, \cdots)$, $\left[\frac{n^{2}}{5}\right]=$ $5 k^{2}+4 k=k(5 k+4)$, it is always a composite number. When $n=5 k+3(k=0,1,2, \cdots)$, $\left[\frac{n^{2}}{5}\right]=5 k^{2}+$ $6 k+1=(k+1)(5 k+1)$, it is always a composite number. When $n=5 k+4(k=0,1,2, \cdots)$, $\left[\frac{n^{2}}{5}\right]=$ $5 k^{2}+8 k+3=(k+1)(5 k+3)$, only when $k=0$, i.e., $n=4$, $\left[\frac{n^{2}}{5}\right]$ is a prime number. In summary, the required sum is $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}$.