Answer: 14

Small Hint:

$\lfloor \log_2(\frac{1}{x}) \rfloor = 2k$ exactly when $2^{-2k-1} \lt x \le 2^{-2k}$

Big Hint:

The region is a product of a set of $x$-intervals and a set of $y$-intervals; sum each set of lengths as a geometric series and multiply

Solution:

For $0 \lt x \le 1,$ the condition $\lfloor \log_2(\frac{1}{x}) \rfloor = 2k$ (for an integer $k \ge 0$) means $2k \le \log_2(\frac{1}{x}) \lt 2k + 1,$ i.e. $x \in \left(2^{-2k-1}, 2^{-2k}\right].$ These intervals have total length $$\sum_{k \ge 0} 2^{-2k-1} = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{2}{3}.$$ Similarly, $\lfloor \log_5(\frac{1}{y}) \rfloor$ is even for $y \in \left(5^{-2k-1}, 5^{-2k}\right],$ intervals of total length $\sum_{k \ge 0} \frac{4}{5} \cdot 25^{-k} = \frac{\frac{4}{5}}{1 - \frac{1}{25}} = \frac{5}{6}.$

The graph of $\mathcal{S}$ is the product of these two sets, so its area is $\frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},$ and $m + n = 5 + 9 = 14.$