Made by me, Inspired from other

Let us consider the last digit of a power of 9.

$\begin{aligned} 9^k & \equiv (10-1)^k \text{ (mod 10)} & \small \color{#3D99F6} \text{where }k \text{ is a non-negative integer.} \\ & \equiv (-1)^k \text{ (mod 10)} \end{aligned}$

$\implies 9^k \equiv \begin{cases} -1 \equiv 9 \text{ (mod 10)} & \text{if } k \text{ is odd.} \\ 1 \text{ (mod 10)} & \text{if } k \text{ is even.} \end{cases}$

Therefore, when $m$ is odd and $n$ is even or $m$ is even and $n$ is odd, $9^m + 9^n \equiv 1 + 9 \equiv 10 \equiv 0 \text{ (mod 10)} \equiv 0 \text{ (mod 5)}$ . Since there are 50 odd integers and 51 even integers from 0 to 100, both inclusive, the number of pairs of $(m, n)$ is $2\times 51 \times 50 = \boxed{5100}$ .