Simple Perfect Squares

Let $n^2 + 45 = k^2 \Rightarrow 45 = (k+n)(k-n).$ Since $45 = 3^{2}5^1$ , there are 6 positive integer divisors to check against: [1, 3, 5, 9, 15, 45]. We now check each of the following integral products:

$k+n = 45; k-n = 1 \Rightarrow n = 22, k = 23$

$k+n = 15; k-n = 3 \Rightarrow n =6, k = 9$

$k+n = 9; k-n = 5 \Rightarrow n = 2, k = 7$

Hence the required values of $n$ are $n = 2, 6, 22$ , which sums to $\boxed{30}.$