Pigeonhole principle problem 104

Question

How can I show that if 19 distinct integers are chosen from the sequence 1, 4, 7, 10, 13, 16, 19 . . ., 97, 100, there must be two of them whose sum is 104. What evidence is there? I am having a bit of trouble solving this. It would be appreciated if you could help. Thanks :)

Answer 1

Find pairs that sum to $104$. You can only have one of each of those pairs. How many pairs are there? How many numbers that are not part of a pair?

Answer 2

Now there are $16$ such pairs which sum to $104$, from $4+100$ to $49+55$, which means that there are a total of $32$ integers in this pairs. In total, there are $\frac{100-1}{3}+1=34$ integers in the sequence, meaning that there are $34-32=2$ integers not in such pairs. This means that if you choose $2+16+1=19$ integers, even assuming that each integer is chosen from each separate pair or not in a pair at all, there still must be at least two of them which sum to $104$.