Solution:

$\mathcal{T}$ is the triangle with vertices $(1,0,0),$ $(0,1,0),$ $(0,0,1).$ Because $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1 = x + y + z,$ whenever two of the inequalities $x \ge \frac{1}{2},$ $y \ge \frac{1}{3},$ $z \ge \frac{1}{6}$ hold, the third can hold only on a boundary segment of zero area. So $\mathcal{S}$ is, up to measure zero, the union of the three regions where a specific pair of inequalities holds.

The region with $x \ge \frac{1}{2}$ and $y \ge \frac{1}{3}$ becomes, after substituting $x = \frac{1}{2} + x'$ and $y = \frac{1}{3} + y',$ a copy of $\mathcal{T}$ with coordinate sum $1 - \frac{1}{2} - \frac{1}{3} = \frac{1}{6},$ i.e. a triangle similar to $\mathcal{T}$ with ratio $\frac{1}{6}$ and area $\left(\frac{1}{6}\right)^2$ of $\mathcal{T}.$ Likewise the pairs $\{x, z\}$ and $\{y, z\}$ give similar triangles with ratios $\frac{1}{3}$ and $\frac{1}{2}.$

The ratio of areas is $$\frac{1}{36} + \frac{1}{9} + \frac{1}{4} = \frac{1 + 4 + 9}{36} = \frac{7}{18},$$ so $m + n = 7 + 18 = 25.$