Consider all your $2n$ points, blue and red, as point charges, each of them having a charge $q_k=\pm1$ and sitting at position $\vec{x}_k$, blue points having charge $+1$ and red points $-1$. Your problem amounts at proving that the following "potential energy" $U$ is always non negative: $$ U(\vec{x}_1,\ldots,\vec{x}_{2n})=-\sum_{i,j; i\ne j}q_iq_j|\vec{x}_i-\vec{x}_j|. $$$$ U(\vec{x}_1,\ldots,\vec{x}_{2n})=-\sum_{\stackrel{\scriptstyle i,j}{i<j}} q_iq_j|\vec{x}_i-\vec{x}_j|. $$ The force derived from such potential is peculiar: two points of the same colour repel each other with unit force directed along the straight line joining them, while points of different colour attract each other in the same way, irrespective of their distance.
Potential energy has a minimum when the total force $\vec{F}_{tot}$ vanishes. The only way to arrange our charges so that $\vec{F}_{tot}=0$ is when every blue point is coupled with a red point, both sharing the same position. But then $U=0$, so that is the minimum of the potential energy.
I don't know if all this can be turned into a rigorous proof, but I hope this approach may give some further insight.