Solution:

The four faces have side multiset $\{5\sqrt{10} \times 8,\ 10 \times 2,\ 18 \times 2\},$ and each edge lies on two faces, so the tetrahedron $ABCD$ has $AB = 10$ and $CD = 18$ as opposite edges and the other four edges equal to $5\sqrt{10}.$ Place $$A = (-5, 0, 12), \quad B = (5, 0, 12), \quad C = (0, -9, 0), \quad D = (0, 9, 0),$$ which is consistent since $AC^2 = 25 + 81 + 144 = 250 = (5\sqrt{10})^2.$ The configuration is symmetric under $x \to -x$ and under $y \to -y,$ so both centers lie on the $z$-axis.

For $S = (0, 0, s),$ equating distances to $A$ and $C$ gives $25 + (12 - s)^2 = 81 + s^2,$ so $s = \frac{11}{3}.$ For $R = (0, 0, t),$ face $ABC$ has plane $4y - 3z + 36 = 0$ and face $ACD$ has plane $12x + 5z = 0,$ so equal distances require $$\frac{36 - 3t}{5} = \frac{5t}{13} \implies t = \frac{117}{16},$$ and by the two mirror symmetries this point is equidistant (at distance $\frac{45}{16}$) from all four faces.

Therefore $RS = \frac{117}{16} - \frac{11}{3} = \frac{351 - 176}{48} = \frac{175}{48},$ which is in lowest terms, so $m + n = 175 + 48 = 223.$