If the boxes hold $a$ and $b$ marbles, then $ab$ is divisible by $50,$ since the black-black probability has denominator $50$

With $a + b = 25,$ only $\{5, 20\}$ and $\{10, 15\}$ work; find black counts whose product is $\frac{27ab}{50}$

Say the boxes hold $a$ and $b$ marbles with $a + b = 25,$ containing $p$ and $q$ black marbles. Then $\frac{pq}{ab} = \frac{27}{50},$ so $50pq = 27ab,$ and since $\gcd(27, 50) = 1,$ we need $ab$ to be divisible by $50.$ Checking $a(25 - a)$ for $a = 1, \ldots, 12,$ only $\{a, b\} = \{20, 5\}$ and $\{10, 15\}$ give a multiple of $50.$

For sizes $20$ and $5:$ $pq = \frac{27 \cdot 100}{50} = 54,$ and since each box also holds a white marble, $p \le 19$ and $q \le 4,$ forcing $p = 18,$ $q = 3.$ The white counts are $2$ and $2,$ so the white-white probability is $\frac{2}{20} \cdot \frac{2}{5} = \frac{1}{25}.$ For sizes $10$ and $15:$ $pq = \frac{27 \cdot 150}{50} = 81,$ and $p \le 9,$ $q \le 14$ force $p = q = 9.$ The white counts are $1$ and $6,$ giving $\frac{1}{10} \cdot \frac{6}{15} = \frac{1}{25}$ again.

Either way the probability is $\frac{1}{25},$ so $m + n = 1 + 25 = 26.$