Mathematical Enigma - X

We require that

$n^{2} + n + 41 = m^{2} \Longrightarrow (n + \frac{1}{2})^{2} - \frac{1}{4} + 41 = m^{2}$

$\Longrightarrow (2n + 1)^{2} - 1 + 164 = 4m^{2} \Longrightarrow (2m)^{2} - (2n + 1)^{2} = 163$

$\Longrightarrow (2m + (2n + 1))(2m - (2n + 1)) = 163.$

Now $163$ is prime, so we have $4$ cases to examine for $(2m + (2n + 1), 2m - (2n + 1)),$ namely $(1,163), (-1,-163), (163,1), (-163,-1).$

If $(2m + (2n + 1), 2m - (2n + 1)) = (1,163)$ or $(-163,-1)$ then upon subtracting the second from the first equation we find that $2*(2n + 1) = -162 \Longrightarrow n = -41.$

If $(2m + (2n + 1), 2m - (2n + 1)) = (-1,-162)$ or $(163,1)$ then upon subtracting the second from the first equation we find that $2*(2n + 1) = 162 \Longrightarrow n = 40.$

The sum of these two possible solutions is then $-41 + 40 = \boxed{-1}.$