You're a double negative

The general answer is that $N$ is the product of all positive primes $p$ such that $(p-1)|L$ .

To see this, note that if $(p-1)|L$ , then either $p|mn$ or $m^L \equiv n^L \equiv 1$ mod $p$ , so all such primes $p$ must divide $N$ .

Also note that $p^2 \nmid N$ for any prime $p$ , because we can take $m = p, n = 1$ .

Finally note that if $(p-1) \nmid L$ , then let $g$ be a primitive root mod $p$ , take $m = g, n = 1$ , and we see that $p \nmid mn(m^L-n^L)$ .

For $L = 80$ we get $2 \cdot 3 \cdot 5 \cdot 11 \cdot 17 \cdot 41 = \fbox{230010}$ .