Let ${A_1, ..., A_n}$ and ${B_1, ..., B_n}$ be two given sets of points in a plane with different centroids. Prove that there exists a point $P$ in the plane such that $\sum |PA_i| = \sum |PB_i|$.
I have solved the problem if we replace equal sum of distances with equal sum of squares of distances. In that case, everything works out nicely using the distance formula and we find that the locus of all such points $P$ is a line perpendicular to the line joining the centroids of ${A_i}$ and ${B_i}$ (call it $G_1G_2$)
The difficulty with the original formulation is that we are summing terms of the form $\sqrt{x_i^2+y_i^2}$, and we need to find a way to get rid of the square roots.
I tried following: Let $P$ vary along the line $l$ joining the centroids $G_1G_2$. First assume $P$ is on the opposite side of $G1$ as $G2$. Let $P$ move very far away from $G1$. We expect that $\sum |PA_i|$ is smaller than $\sum |PB_i|$. Note that this is obvious if the convex hull of $A_i$ and the convex hull of $B_i$ are disjoint, but we need to prove it in the general case.
So the difference $\sum |PA_i|-\sum |PB_i|$ is negative. By the same argument if $P$ is on the opposite side of $G2$ as $G1$, then as it moves far away the difference $\sum |PA_i|-\sum |PB_i|$ is positive. Since the difference is continous, it must be zero for some point $P$ on $G_11G_2$
Since the centroids are different, we can pick a rectangular coordinate system in which the centroids have different $x$-coordinates.
Now write $A_i = (a_i,a_i')$ and $B_i = (b_i,b_i')$. The hypothesis implies that $\sum a_i \ne \sum b_i$. Let $f(x) = \sum |PA_i| - \sum |PB_i|$ for $P = (x,0)$.
We have $|PA_i| = \sqrt{(x-a_i)^2 + a_i'^2} = |x|\sqrt{1 - 2a_i/x + (a_i^2 + a_i'^2)/x^2} = |x| - a_i \operatorname{sgn}(x) + o(1)$ as $x \to \pm \infty$.
Thus $f(x) \to \operatorname{sgn}(x)(\sum b_i - \sum a_i) \ne 0$ as $x \to \pm \infty$. Therefore $f(x)$ takes both positive and negative values. Since $f$ is continuous, it must take the value zero for some choice of $x$.