For any vertex
x, count the number
n(x) of time any edge adjacent to
x is taken. It must be an even number since it's a circuit (you have to enter the vertex and then leave it each time you visit it). Now if
x is among the
2m odd degree vertex, one of the adjacent edge must have been taken twice. Indeed, all the adjacent edge are taken, their number is odd, and there has been an even number of adjacent edges taken. As a consequence, for any vertex of degree 2m, there is at least an edge adjacent to it which appear twice in the circuit. Each edge is adjacent to less that two vertices (maybe one if you allow loops), so that you need at least m of them to be adjacent to the 2m vertices of odd degree.