= Solution
Fix any seven-element <subset> $S$. Its <power set> supplies $2^7=128$ pigeons. Map each <subset> to its <integer> sum. Since the elements are positive, all such sums lie between $0$ and $\sum_{s\in S}s$. The latter is at most the sum of the seven largest available elements,
$$
10+11+\cdots+16=91.
$$
There are therefore at most $92$ possible sums, fewer than the $128$ <subsets>. The <pigeonhole principle> forces two distinct <subsets> to have the same sum. \b[Every seven-element subset has the two-sum property.] This is the <subset-sum pigeonhole bound>. The weaker estimate $\sum_{s\in S}s\leq7\cdot16=112$ would also suffice, since $128>113$.
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