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?

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?

Consider starting with a graph $G$ and partitioning its vertices into $V_1$ and $V_2$ by looking at each vertex one by one and assigning it to $V_1$ or $V_2$ (starting with $V_1$ and $V_2$ both empty): if the vertex $v$ has more edges going to $V_1$ than $V_2$, assign it to $V_2$, otherwise assign it to $V_1$. If you assign $v$ to $V_i$, colour each edge from $v$ to $V_i$ red and every edge from $v$ to $V_{3-i}$ blue. Then there will always be at least as many blue edges as red edges. When the process is finished, all edges will be coloured, those within $V_1$ or $V_2$ will be red and those between $V_1$ and $V_2$ will be blue.

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.