This is tricky...

Since $100 = 1 + (34 - 1)*3$ we know that there are $34$ terms in the A.P..

Now the $19$ th term is $1 + (19 - 1)*3 = 55$ . If $A$ were the set of the first $19$ terms of the A.P. then the largest possible sum would be $55 + 52 = 107$ , and the next largest would be $55 + 49 = 104$ . So $104$ is the only possible answer of the options given.

Now we need to show that for any $A$ there will be two elements that sum to $104$ . Now two terms $a_{m} = 1 + (m - 1)*3$ and $a_{n} = 1 + (n - 1)*3$ will sum to $104$ if

$a_{m} + a_{n} = 2 + 3*(m + n - 2) = 3*(m + n) - 4 = 104 \Longrightarrow m + n = 36$ .

Without loss of generality let $m \lt n$ .Now since $m, n \le 34$ and must be distinct, there are $16$ possible pairs $(m,n)$ such that $m + n = 36$ . These are $(2, 34), (3, 33), (4, 32), ....... , (16, 20), (17, 19)$ . The only values that cannot be part of a pair are $1$ and $18$ . So we can choose one value from each of the $16$ pairs along with $1$ and $18$ and not have a pair of elements that add to $36$ , but the next value we choose will necessarily "complete" one of the $16$ pairs, (by the Pigeonhole Principle).

Thus for any set of $19$ distinct integers chosen from the given A.P. there must be a pair of elements that add to $104$ .