Perfect Square

Perfect squares are equivalent to either $0 \pmod{4}$ or $1 \pmod{4}$ .

For $N \ge 4$ we have that $N! + 115 \equiv 3 \pmod{4}$ , so we are looking for $N \lt 4$ .

Since $1! + 115 = 116, 2! + 115 = 117$ and $3! + 115 = 121 = 11^{2}$ , we can conclude that there is only $\boxed{1}$ natural number $N$ , namely $N = 3$ , such that $N! + 115$ is a perfect square.