Solution:

Note that the remainder when $n^{2}$ is divided by $5$ must be $0$, $1$, or $4$. Then we have that $25p = n^{2}$ or $25p = n^{2} - 1$ or $25p = n^{2} - 4$. In the first case there are no solutions. In the second case, if $25p = (n-1)(n+1)$, then we must have $n-1 = 25$ or $n+1 = 25$ as $n-1$ and $n+1$ cannot both be divisible by $5$, and also cannot both have a factor besides $25$. Similarly, in the third case, $25p = (n-2)(n+2)$, so we must have $n-2 = 25$ or $n+2 = 25$.

Therefore the $n$ we have to check are $23$, $24$, $26$, $27$. These give values of $p = 21$, $p = 23$, $p = 27$, and $p = 29$, of which only $23$ and $29$ are prime, so the answer is $23 + 29 = 52$.