Assume that 2m red points and 2n blue points are given on the lattice Z² in the plane R². We show that if they are in general position, that is, if at most one point lies on each vertical line and horizontal line, then there exists a rectangular cut that bisects both red points and blue points. Moreover, if they are not in general position, namely if some vertical and horizontal lines may contain more than one point, then there exists a semi-rectangular cut that bisects both red points and blue points. We also show that these results are best possible in some sense. Moreover, our proof gives O(N log N), N = 2m + 2n, time algorithm for finding the desired cut.