1. The sum of a set of numbers is the sum of all its elements. Let $S$ be a set of positive integers not exceeding 15, such that the sums of any two disjoint subsets of $S$ are not equal. Moreover, among all sets with this property, the sum of $S$ is the largest. Find the sum of the set $S$.
(4th American Invitational Mathematics Examination)
1. First, prove that $S$ has at most 5 elements. In fact, if $S$ has at least 6 elements, then the number of non-empty subsets of $S$ with at most 4 elements is at least $C_{6}^{1} + C_{6}^{2} + C_{6}^{3} + C_{6}^{4} = 56$ (subsets). The sum of these subsets is no more than 54 (i.e., $15 + 14 + 13 + 12 = 54$). By the pigeonhole principle, among these 56 numbers that do not exceed 54, at least two must be equal, meaning that at least two subsets have the same sum. If these two subsets are disjoint, it contradicts the condition that "the sums of any two disjoint subsets of $S$ are not equal." If these two subsets intersect, removing the common elements still leads to a contradiction. Therefore, $S$ has at most 5 elements.

Next, construct this 5-element set to maximize the elements (and thus the sum of the set). After $S$ contains $15, 14, 13$, it cannot contain 12, otherwise, $15 + 12 = 14 + 13$. $S$ can contain 11, but not 10 or 9, otherwise, $15 + 10 = 14 + 11$, $15 + 9 = 11 + 13$. The last element is 8, so $S = \{15, 14, 13, 11, 8\}$, and $15 + 14 + 13 + 11 + 8 = 61$. Therefore, the largest possible sum is 61.