Geometry?

Let $a_i = \arctan {x_i}$ . Notice that if $|a_i - a_j| \leq \frac{\pi}{6}$ , then $|\frac{x_i-x_j}{1+x_ix_j}| = | \tan(a_i - a_j)| \leq \frac{1}{\sqrt{3}}$ . Notice also that since $0 \leq a_i \leq \pi$ , we must have at least one pair $(i,j)$ for which $|a_i - a_j| \leq \frac{\pi}{6}$ . But then, if $|a_i - a_j| \geq \frac{\pi}{6}$ , this implies that we equally distributed the $a_i$ 's along the interval $[0, \pi]$ , implying one of the $a_i$ 's equal $\frac{\pi}{2}$ , which implies $x_i$ is undefined, which is not possible. As such, there must always be $i,j$ for which $|a_i - a_j| < \frac{\pi}{6}$ , implying that $|\frac{x_i-x_j}{1+x_ix_j}| < \frac{1}{\sqrt{3}}$ . Thus, $D= \frac{1}{\sqrt{3}}$ , and equality cannot be attained. This implies that our answer is $\frac{1}{D^2} + 1 = \sqrt{3}^2 + 1 = \boxed{4}$ .