12. The sum of a set of numbers is the sum of all its elements. Let $S$ be a set composed of positive integers not exceeding 15, such that the sums of any two disjoint subsets of $S$ are not equal, and among all sets with the above property, the sum of $S$ is the largest. Find the set $S$ and its sum.
12. First, $S$ can contain at most 5 elements, because otherwise, $S$ would have at least 6 elements, and the number of subsets with elements not exceeding 4 would be at least $C_{8}^{1} + C_{8}^{2} + C_{8}^{3} + C_{8}^{4} = 56$. The sum of each such subset is at most 54 (since $15 + 14 + 13 + 12 = 54$). By the pigeonhole principle, at least two subsets would 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 they intersect, removing their common elements would still lead to a contradiction. Therefore, $S$ can contain at most 5 elements.

Next, construct a 5-element subset with the largest possible elements. After including $15, 14, 13$, $S$ cannot include 12 (since $15 + 12 = 14 + 13$). $S$ can include 11. If $S$ includes $15, 14, 13, 11$, it cannot include 10 or 9 (since $10 + 15 = 14 + 11$ and $9 + 15 = 13 + 11$). Therefore, the fifth element of $S$ can only be 8. It is not difficult to verify that $S = \{8, 11, 13, 14, 15\}$ satisfies the problem's conditions, so we conjecture that $8 + 11 + 13 + 14 + 15 = 61$ is the solution.

Finally, prove that the maximum value is indeed 61. If a set $S$ satisfying the conditions can be constructed with a sum greater than 61, then $S$ must include $15, 14, 13$ (because even if it does not include the smallest of these, 13, to exceed a sum of 61, it would need to include $10, 11, 12$, which would result in $15 + 11 = 14 + 12$, a contradiction). Thus, $S$ cannot include 12. If $S$ includes 11, it would lead to the set with a sum of 61, a contradiction. If $S$ does not include 11, even if $S$ includes 10 and 9, it can only make the sum reach $15 + 14 + 13 + 10 + 9 = 61$, a contradiction. Therefore, 61 is indeed the maximum value.
Thus, the sum of the set $S$ is 61.