T For Two

Group the terms in $S$ as follows: $(1,103),(199,7),\ldots,(109,97)$ . There are totally 17 groups.

If 2 elements are chosen from the same group other than the first group, then will sum up to 206.

By pigeonhole principle , the minimum number to be chosen to guarantee the sum of 206 is $17+1+1=19$

Note that we added an extra 1 as we can choose both the elements from the first group.

The elements of $S$ are of the form $6k + 1$ for $k = 0,1,2,3, ..., 33$ . We are required to find two elements $6m + 1$ and $6n + 1$ , where $m, n$ have the same domain as $k$ with $m \ne n$ , such that

$6m + 1 + 6n + 1 = 206 \Longrightarrow 6(m + n) = 204 \Longrightarrow m + n = 34$ .

Since $m \ne n$ we can't have $m = n = 17$ , and neither of $m,n$ can be $0$ as the other element would need to be $34$ , which is not in the domain of $k$ . The other $32$ numbers form $16$ pairs that add to $34$ , namely $(1,33), (2,32), .... , (15,19), (16,18)$ . We can then choose one element from each of these $16$ pairs, along with $0$ and $17$ , and still not have two distinct elements that add to $34$ , but by the Pigeonhole Principle, the next element chosen guarantees that we will two distinct elements that do add to $34$ . So $N = 16 + 2 + 1 = \boxed{19}$ .