simple

$m+n+mn=90$ . Therefore, $1+m+n+mn=91$ or $(1+m)(1+n)=91$

91 can be expressed in 2 ways only,namely $91\times1$ and $13\times7$ .

Hence, the non-negative solutions are $(m,n)=(90,0);(0,90);(12,6);(6,12)$

But since, $m$ and $n$ are positive,we cannot have $(90,0)$ and $(0,90)$ as solutions.

In the other two solutions,sum of $m$ and $n$ is $18$ .

$m + n = 90 - mn \\ n = 90 - mn - m \\ n = 90 - m(n+1) \\ m(n+1) = 90 - n \\ m = \frac {90 -n}{n+1} \\ m = \frac {90 -(n-1)}{n+1} \\ m = \frac {91}{n+1} - 1 \\ m +1 = \frac{91}{n+1} \\ (m+1) * (n+1) = 91 = 13*7 \\ m = 12, \quad n = 6, \quad or \quad m = 6, \quad n=12, \quad anyway \quad \\ m+n =18$

MN=72 M+n=18 M=12 N=6 Or M=6 N=12