Question 15
For $1 \le i \le 215$ let $a_i = \frac{1}{2^i}$ and $a_{216} = \frac{1}{2^{215}}.$ Let $x_1,$ $x_2,$ $\ldots,$ $x_{216}$ be positive real numbers such that $$
\sum_{i=1}^{216} x_i = 1
$$ and $$
\begin{aligned} &\sum_{1 \le i \lt j \le 216} x_i x_j \\ &= \frac{107}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{2(1 - a_i)}. \end{aligned}
$$ The maximum possible value of $x_2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
\sum_{i=1}^{216} x_i = 1
$$ and $$
\begin{aligned} &\sum_{1 \le i \lt j \le 216} x_i x_j \\ &= \frac{107}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{2(1 - a_i)}. \end{aligned}
$$ The maximum possible value of $x_2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$