Answer: 318

Small Hint:

A flap folded down from a vertex hangs to a depth equal to the distance from that vertex to its fold line, so all three fold lines must be at distance $h$ from their vertices

Big Hint:

Two folds cutting the same side must not overlap: for each side, multiplying $h$ by the sum of the other two side lengths gives at most twice the triangle’s area; the largest such sum is the binding constraint

Solution:

Write $a = BC = 23,$ $b = CA = 27,$ $c = AB = 30,$ and let $K$ be the area of $\triangle ABC.$ By Heron’s formula with semiperimeter $40,$ $K = \sqrt{40 \cdot 17 \cdot 13 \cdot 10}$ $= 20\sqrt{221}.$ When the corner at a vertex is folded down at a right angle, the flap hangs to a depth equal to the distance from that vertex to the fold line, so for a level tabletop of height $h,$ each fold line must lie at distance $h$ from its vertex.

The flap at $A$ is similar to $\triangle ABC$ with ratio $\frac{h}{\frac{2K}{a}} = \frac{ha}{2K}$ (dividing $h$ by the distance from $A$ to $\overline{BC}$), so it uses up $AU = c \cdot \frac{ha}{2K}$ of side $\overline{AB};$ likewise the flap at $B$ uses $BZ = c \cdot \frac{hb}{2K}$ of the same side. The two folds fit without crossing exactly when $AU + BZ \le c,$ that is, $h(a + b) \le 2K.$ The other two sides give $h(b + c) \le 2K$ and $h(c + a) \le 2K.$

The binding constraint comes from the largest sum, $b + c = 57,$ so the maximum height is $$h = \frac{2K}{57} = \frac{40\sqrt{221}}{57},$$ and $k + m + n = 40 + 221 + 57$ $= 318.$