Solution:

Cut the plane into six $60^{\circ}$ "pizza slices" with vertex $Q$. Rotating if necessary, we may assume that none of the $P_{j}$ lie on the cuts. By the pigeonhole principle, one slice contains at least $n / 6$ of the $P_{j}$. Let $P_{i}$ be a point in this slice farthest from $Q$. It remains to show that all other points $P_{j}$ in this slice satisfy $P_{j} P_{i} < P_{i} Q$.

The average of the angles of $\Delta P_{j} P_{i} Q$ is $180^{\circ} / 3 = 60^{\circ}$, so $\angle P_{j} Q P_{i}$, which is less than $60^{\circ}$, is less than one of the other angles. Smaller angles of a triangle are opposite shorter sides, so $P_{j} P_{i}$ is less than one of $P_{j} Q$ and $P_{i} Q$. By choice of $i$, $P_{j} Q \leq P_{i} Q$, so in any case, $P_{j} P_{i} < P_{i} Q$. (Alternatively, one could use the Law of Cosines to show that the side of a triangle opposite an angle smaller than $60^{\circ}$ is not the longest.)