RHS = Perfect square

$n^2-8n-5=N^2$ where $N$ is a positive integer. Now rewrite $n^2-8n-5$ as $(n-4)^2-21$ .

Thus the original equation changes to $(n-4)^2-21=N^2\implies (n-4)^2-N^2=21$ Factorizing the above equation, we get, $(n-4+N)(n-4-N)=1 \times 3 \times 7$ .

Case 1: $n-4+N=21,\; n-4-N=1$ . Solving this we get $(n,N)=(15, 10)$

Case 2: $n-4+N=7,\; n-4-N=3$ . Solving this we get $(n,N)=(9, 2)$

Note: The remaining cases will just give you the negative values for $N$ which is also correct as $N$ is getting squared and the same values for $n$ .

Thus the sum of all possible values for $n$ is $15+9=\boxed{24}$