Assume 14 < n < 30. Make boxes labelled 16-n to 30-n. Let Si be the set of the first i elements. Let si be the sum of the elements of Si. If si is less than or equal to 30-n, put Si in the box labelled si. Otherwise, put it in the box labelled si-n. There are 15 boxes, and 16 subsets, so at least one box has two subsets. Clearly, a box with two subsets must have one, say Si, with si = the box label, and one subset Sj with sj-n = the box label, with Sj containing Si, obviously. So Sj - Si is a subset whose elements have sum n. The complement of this set with respect to the full set of 16 elements has sum 30-n.
You can probably express the above in 4 lines if you're especially terse.