[Solution] We 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
$$C_{6}^{1}+C_{6}^{2}+C_{6}^{3}+C_{6}^{4}=56 \text { (subsets) }$$

The sums of these subsets do not exceed 54 (i.e., $15+14+13+12=54$). By the pigeonhole principle, among 56 positive integers not exceeding 54, at least two numbers are equal, i.e., 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, we construct this 5-element set to maximize the elements (and thus maximize 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 it cannot contain 10 and 9, otherwise $15+10=14+11, 15+$
$$9=13+11;$$

The last element is 8, then
$$\begin{array}{l} S=\{15,14,13,11,8\} \\ 15+14+13+11+8=61 \end{array}$$

Thus, the largest possible sum is 61.