TITLE: In a graph, the vertices can be partitioned $V=V_1\cup V_2$ so that at most half of all edges run within each part?
QUESTION [6 upvotes]: I am thinking of destroying all cycles of odd length by removing edges, so that I get a bipartite graph, with a partition $V_1$ and $V_2$ so that no edge in the new graph run within the two parts. Then the worst case in adding back the edges I removed, is that these edges all run within each part. 
So I would like to show that I can destroy all odd cycles by removing at most half of all the edges. Anyone can tell me if I am thinking in the right direction?

REPLY [7 votes]: This is a classic problem where the Probabilistic Method gives a very nice solution.
Consider the effect of randomly assigned each vertex, independently and identically, to $V_1$ or $V_2$ with probability $1/2$. In this case, the expected number of edges that cross the two halves is $m/2$. This implies that there is a positive probability of partitioning the vertices so that the actual (as opposed to expected) number of edges crossing the two halves is at least $m/2$. In particular, such a partition exists.