Hello paul14Welcome to Math Help Forum!

I reckon the answer is 23. Here's my reasoning.

Consider the partition of P into sets as follows:

So, for example, , etc

Then and

Now if we choose an to be in our subset , then we must not choose an , for then

Thus the subset can contain at most element from elements from , and then elements from any two of the remaining , in such a way that no two of the 's add up to .

Total:

Grandad