There is a brute $O(n^3)$ algorithm.
There are $\binom n 2 = \frac {n(n-1)} 2$ lines over pairs of points. If a line goes over $k$ points, there are $2(k+1)$ ways to rotate the line by a negligible angle, so that no points go over it and each rotation partitions the $k$ points differently.
For each pair of vertices (v1, v2),
rotate the space so that v1.y = v2.y = 0
let UP be the sum of values of vertices with positive y
let DOWN be the same for negative y
let L be a list of vertices with y = 0, sorted by x
For each of L.length+1 ways to cut L in two,
let LEFT be the sum of values of vertices in L to the left of the cut
let RIGHT be the same for vertices to the right
check whether one of UP + LEFT, DOWN + RIGHT or UP + RIGHT, DOWN + LEFT
is a better partition than your previous best.