Answer: 863

Solution:

Since $\sum x_i = 1,$ we have $2\sum_{i \lt j} x_i x_j = 1 - \sum x_i^2.$ Doubling the given equation and rearranging, $$\begin{aligned} 1 - \sum_{i=1}^{216} x_i^2 &= \frac{214}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{1 - a_i}, \end{aligned}$$ so $$\begin{aligned} \frac{1}{215} &= \sum_{i=1}^{216}\left(1 + \frac{a_i}{1 - a_i}\right)x_i^2 \\ &= \sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}. \end{aligned}$$

Now $\sum a_i = \left(\frac{1}{2} + \cdots + \frac{1}{2^{215}}\right)$ $+ \frac{1}{2^{215}} = 1,$ so $\sum (1 - a_i) = 216 - 1 = 215.$ By the Cauchy-Schwarz inequality, $$\begin{aligned} 1 &= \left(\sum_{i=1}^{216} x_i\right)^2 \\ &\le \left(\sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}\right) \\ &\quad {}\cdot \left(\sum_{i=1}^{216} (1 - a_i)\right) \\ &= \frac{1}{215} \cdot 215 = 1. \end{aligned}$$

Equality holds, so $x_i$ is proportional to $1 - a_i,$ forcing $x_i = \frac{1 - a_i}{215}.$ The only, hence maximum, possible value of $x_2$ is $\frac{1 - \frac{1}{4}}{215} = \frac{3}{860},$ and $m + n = 3 + 860 = 863.$