2016 AIME II Problem 15

For 1 \leq i \leq 215 let a_i = \dfrac{1}{2^{i}} and a_{216} = \dfrac{1}{2^{215}}. Let x_1, x_2, ..., x_{216} be positive real numbers such that \sum_{i=1}^{216} x_i=1 and \sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}. The maximum possible value of x_2=\dfrac{m}{n}, where m and n are relatively prime positive integers. Find m+n.

Leading zeroes must be inputted, so if your answer is 34, then input 034

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.