Let's try to proove it with one additional statement: "There is only one type of diploms with some size S".
First we put all type of diploms in seperate lines. Now we must minimize the number of rows. By the problem description we have: We can merge two (x and y) rows into one if and only if equality is statisfied: |size(x)-size(y)| == 1
So, some diplom row x can be merged only with row y. Where: size(y) == size(x)+1 or size(y) == size(x)-1.
If none of y diploms row exist, then row x can not be merged and it must stay.
If only one dimplom row y exist, then we must do this megre not looking on y at all. Even if y has other possible merging global minimus isn't affected. If we can megre x and y or y and z, we can merge only one pair.
The last type of merge left, then x can be merged from both sides. Here we cant choose the prefered merge so easily. So let's sort all rows and analise them in accessing order. Then this situation just can not happen, cause all size(y)-1 merging row will be merged before, then analysing Y row.
Now lets apply this proof to out problem. We just have to divide all dimploms into sets. In set A goes all dimpols x with size(x) == A. So we must minimize the sum of all set's power by same apporach.