6. All natural numbers $n$ that make $\left[\frac{n^{2}}{5}\right]$ a prime number, the sum of their reciprocals is ( )
A. $\frac{9}{20}$
B. $\frac{11}{30}$
C. $\frac{5}{12}$
D. $\frac{37}{60}$
6. D Let $m=\left[\frac{n^{2}}{5}\right]$.
(1) When $n=5 k\left(k \in \mathbf{N}_{+}\right)$, $m=\left[\frac{25 k^{2}}{5}\right]=5 k^{2}$, only when $k=1$, $m$ is a prime number, at this time $n=5$.
(2) When $n=5 k+1(k \in \mathbf{N})$, $m=\left[\frac{(5 k+1)^{2}}{5}\right]=5 k^{2}+2 k=k(5 k+2)$, only when $k=1$, $m$ is a prime number, at this time $n=6$.
(3) When $n=5 k+2(k \in \mathbf{N})$, $m=\left[\frac{(5 k+2)^{2}}{5}\right]=5 k^{2}+4 k=k(5 k+4)$. When $k=1$, $m=9$ is a composite number, therefore for all $k \in \mathbf{N}_{+}, m$ is a composite number.
(4) When $n=5 k+3(k \in \mathbf{N})$, $m=\left[\frac{(5 k+3)^{2}}{5}\right]=(5 k+1)(k+1)$. When $k=0$, $m=1$, when $k \in \mathbf{N}_{+}$, $m$ is a composite number.
(5) When $n=5 k+4(k \in \mathbf{N})$, $m=\left[\frac{(5 k+4)^{2}}{5}\right]=(5 k+3)(k+1)$. 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, thus the sum of the reciprocals of $n$ is $\frac{37}{60}$.