The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}.$ The sum of all possible values of n is ____________

Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater). Break up the problem into two cases: an even number of sides 2n, or an odd number of sides 2n – 1. For polygons with 2n sides, the circumdiameter has endpoints on 2 vertices. There are n – 1 points on one side of a diameter, pulse 1 of the endpoints of the diameter for a total of n points. For polygons with 2n -1 points, the circumdiameter has 1 endpoint on a vertex and 1 endpoint on the midpoint of the opposite side. There are also n – 1 points on one side of the diameter, plus the vertex for a total of n points on one side of the diameter. 

Case 1 : 2n – sides polygon. There are clearly $\left(\begin{array}{c}2n\\ 3\end{array}\right)$ different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously 2n choices for this point. From there, the other two points must be within the n – 1 points remaining on the same side of the diameter. So our desired probability is $$\begin{gathered} \frac{2n\left(\begin{array}{c}n-1\\ 2\end{array}\right)}{\left(\begin{array}{c}2n\\ 3\end{array}\right)}=\frac{n(n-1)(n-2)}{\frac{2n(2n-1)(2n-2)}{6}}=\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}=\frac{3(n-2)}{2(2n-1)} \\ so \frac{93}{125}=\frac{3(n-2)}{2(2n-1)}. \end{gathered}$$

$$\begin{gathered} 186(2n-1)=375(n-2), 372n-186=375n-750 \\ 3n=564 \\ n=188 and so the polygon has 376 sides. \end{gathered}$$

Case 2 :  2n -1-sided polygon. Similarly, $\left(\begin{array}{c}2n-1\\ 3\end{array}\right)$ total triangles. Again choose the leftmost point, with 2n -1 choices. For the other two points, there are again $\left(\begin{array}{c}n-1\\ 2\end{array}\right)$ possibilities. The probability is 

$$\begin{gathered} \frac{(2n-1)\left(\begin{array}{c}n-1\\ 2\end{array}\right)}{\left(\begin{array}{c}2n-1\\ 3\end{array}\right)}=\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}=\frac{3(n-2)}{2(2n-3)} so \frac{93}{125}=\frac{3(n-2)}{2(2n-3)} \\ 186(2n-3)=375(n-2) \\ 375n-750=372n-558 \\ 3n=192 \\ n=64 and our polygon has 127 sides. Adding. 127 + 376 = \boxed{503} \end{gathered}$$