Answer: 210

Small Hint:

There are $7$ pairs summing to $16$ and $8$ pairs summing to $24;$ split by whether the two pairs can be chosen disjoint

Big Hint:

If the pairs must overlap, some element $a$ has both $16 - a$ and $24 - a$ in the subset plus a free fourth element; subtract the subsets that have two such centers

Solution:

The pairs of distinct elements summing to $16$ are $\{1,15\}, \{2,14\}, \ldots, \{7,9\}$ (seven pairs), and those summing to $24$ are $\{4,20\}, \{5,19\}, \ldots, \{11,13\}$ (eight pairs). First count subsets containing a $16$-pair and a $24$-pair that are disjoint. Of the $7 \cdot 8 = 56$ combinations, the ones sharing an element $x$ require $16 - x,$ $x,$ and $24 - x$ all to be valid, which happens for the $10$ values $x \in \{4, \ldots, 15\}$ other than $8$ and $12.$ No four-element set arises from two different disjoint combinations (a second decomposition would force a $16$-pair to coincide with a $24$-pair), so this case gives $56 - 10 = 46$ subsets.

In the remaining subsets every $16$-pair meets every $24$-pair, so some center $a$ has both $b = 16 - a$ and $c = 24 - a$ in the subset. There are $10$ possible centers ($a \in \{4, \ldots, 15\}$ with $a \ne 8, 12$), and the fourth element can be any of the $17$ remaining numbers, giving $170$ center–subset counts. Exactly $6$ subsets admit two centers and are counted twice: $\{1,7,9,15\},$ $\{2,6,10,14\},$ $\{3,5,11,13\},$ $\{5,11,13,19\},$ $\{6,10,14,18\},$ and $\{7,9,15,17\}.$ This case gives $170 - 6 = 164$ subsets, none of which contain disjoint pairs.

The total is $46 + 164 = 210.$