2011 AIME I Problem 10

A triangle inscribed in a circle is obtuse exactly when its three vertices lie strictly inside some semicircle

Count obtuse triangles by their clockwise-first vertex, and handle even and odd $n$ separately (only even $n$ has right triangles)

By the inscribed angle theorem, an inscribed triangle is obtuse exactly when its three vertices lie strictly within some semicircle. Count obtuse triangles by their “first” vertex, the vertex from which the other two are reached going clockwise within half the circle. If $n = 2k,$ the open semicircle clockwise of a vertex contains $k - 1$ vertices, giving $n\binom{k-1}{2}$ obtuse triangles; if $n = 2k + 1,$ it contains $k$ vertices, giving $n\binom{k}{2}.$

For $n = 2k$ the probability is $$\frac{2k\binom{k-1}{2}}{\binom{2k}{3}} = \frac{3(k - 2)}{2(2k - 1)} = \frac{93}{125},$$ so $375(k - 2) = 186(2k - 1),$ giving $3k = 564,$ $k = 188,$ and $n = 376.$

For $n = 2k + 1$ the probability is $\frac{3(k - 1)}{2(2k - 1)} = \frac{93}{125},$ so $375(k - 1) = 186(2k - 1),$ giving $3k = 189,$ $k = 63,$ and $n = 127.$ The sum of all possible values is $376 + 127 = 503.$