Answer: 231

Solution:

Since $20^{20} = 2^{40} \cdot 5^{20},$ a valid $m$ must have the form $2^a 5^b,$ and then $n = 2^{40 - 2a}\,5^{20 - 2b}$ is a positive integer exactly when $2a \le 40$ and $2b \le 20.$ Conversely every such choice works, and $n$ is uniquely determined by $m.$

So $a$ can be any of $0, 1, \ldots, 20$ and $b$ any of $0, 1, \ldots, 10,$ giving $21 \cdot 11 = 231$ ordered pairs.