In all cases, the total number of sums is the number of combinations of m elements out of n elements which is mCn.

You may choose m even numbers and 0 odd numbers to the total of mCa·0Cb, or you may choose (m-2) even numbers and 2 odd numbers to the total of (m-2)Ca·2Cb, or (m-4) even numbers and 4 odd numbers to the total of (m-4)Ca·4Cb, and so on. You'll have to add up all the products and divide the result by mCn. Can't tell you off the top of my hat whether the result may be given by a simple formula. (I assumed rCs = 0 wherever r > s.)