Each operation changes the amount of red and blue by 2. WLOG assume the number of red edges is $2k$. Then, we must have at least k operations involving two red edges. Let (ABC) denote the operation on points A, B, and C Consider a group of four vertices. Let's see what we can do with them. 1 red edge: We can "move" this edge to any other edge in the group, which still keeping a total of one red edge within the group. For example, if we have the quadrilateral ABCD with only AB red, we can do (ABC), (ABD), (ACD), (ABD) to get BC to become red. 2 red edges: It's easy to see we can always change these two edges to blue with some operations 3 red edges: Perform one operation to change it to only one red edge 4 red edges: We can always remove all of these 5 red edges: Perform two operations to change it to only one red edge 6 red edges: Choose another quadrilateral. Since this never increases, and we can move lone edges around, we will always be able to convert all of the red edges into blue (Also, there must be at least one quadrilateral in the beginning with less than 6 red edges).

Induction on number of vertices : (Using the operation we can monochrome a Kn(complete graph with n vertices) with blue and red edges) Obviously its true for n=3. For n=k : Consider a vertic A: if all edges from A was wlog blue obviously we can make entire graph blue. if not by assumption of induction we can make entire graph excluding A monochromatic ( wlog blue ). then by applying the operation on any pair of blue and red edges getting out from A and the other edge which connects them all reds will become blue and now entire graph is blue. $Q.E.D.$