Japan Junior Mathematical Olympiad

From $m \neq n$ and the fact that $n$ is a multiple of $m$, it follows that $n \geq 2m$. Since $m$ is an integer greater than or equal to $100$, we have $n - m \geq m \geq 100$, and therefore, we see that $n$ and $m$ differ on the hundred's digit. This means that there exists an integer $k$ ($1 \leq k \leq 8$) for which $n - m = 100k$. Since $m$ divides $n$, we see that $m$ divides $n - m = 100k$. Also, from $2m \leq n \leq 999$ it follows that $m \leq 499$. From these considerations we conclude that $m$ must be one of the following:

$100$, $120$, $125$, $140$, $150$, $160$, $175$, $200$, $250$, $300$, $350$, $400$.

Values of $n$ matching these values of $m$ can be chosen for $8$, $1$, $1$, $1$, $2$, $1$, $1$, $3$, $1$, $2$, $0$, $1$ different ways, respectively. Hence the total number of ways the pair $(m, n)$ can be chosen to satisfy the conditions of the problem is $22$.