Proving problem NP-completeness
First, when doing complexity analysis, we usually think in terms of decision problems, i.e. the answer must be yes or no, and here you have an optimisation problem. This is why you are struggling to find a certificate here. If you reformulate your problem as a decision problem with an additional parameter $k$, and the question become "Is there a subset I such that the sum (...) Is greater than k ?", You will easily find the certificate : the list of the vertices in I, you can polynomially check that there are as many elements of V and W, and that the sum has the desired property.

Second, it seems like your are trying to reduce your problem to VERTEX-COVER. However, we already know that's possible because VC is NP-complete. To show that your problem is NP-Hard, you have to reduce a NP hard problem to it. 

Finally, the reduction : maybe something from VERTEX-COVER will work : take an instance with a graph G=(V,E), make it complete by adding edges and put their weight to 0 (then your problem will not "try" to get them). If you find a clever way to divide V into V and W, the reduction might work.
However, it seems to me that VERTEX-COVER does not have a condition on sets and partitions, and you might have more luck with Subset-sum or 2-partition.

Edit : after a bit of thinking, If you make 2 copies of G (one becomes your V and the other W), and complete it with 0-weighted edges, you juste have to find the correct value for k... I think that's not too far.

I hope that helps.