We could take any element from 65 to 100 without violating the given condition as they cannot add to any positive integer to give 65. i.e. 36 elements
Now, we can get 65 by $(1+64),(2+63) \dots (32+33)$.
So, for each of the above pair, either take an element from $\{1,2,3,\dots,32\}$ or from $\{33,34,\dots 64\}$ => i.e. we get 32 elements.
So, we can have up to $32+36 = 68$ elements without 2 elements adding to 65.