Example 3.3.3 Let $S$ be a set composed of some positive integers not greater than 15, and any two disjoint subsets of $S$ have unequal sums of elements. Find the maximum value of the sum of elements in $S$ with this property.
Solution: If $|S|=n \geqslant 6$, we prove that $S$ does not satisfy the problem's conditions. The number of non-empty subsets of $S$ with no more than 4 elements is
$$
\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{4} \geqslant \mathrm{C}_{6}^{1}+\mathrm{C}_{6}^{2}+\mathrm{C}_{6}^{3}+\mathrm{C}_{6}^{4}=56
$$

Since $12+13+14+15=54$, it is evident that the sum of elements in each of the above subsets does not exceed 54. Given $54<56$ and by the pigeonhole principle, there must be at least two subsets $A$ and $B$ whose elements have the same sum. If $A \cap B=\varnothing$, this contradicts the problem's conditions. If $A \cap B \neq \varnothing$, then replacing $A, B$ with $A \backslash(A \cap B)$ and $B \backslash(A \cap B)$ still leads to a contradiction. Therefore, to satisfy the problem's conditions, we must have $|S| \leqslant 5$.
Next, we construct a 5-element subset $S$ that satisfies the problem's conditions and has the maximum possible sum of elements.
To maximize the sum of elements in $S$, the elements of $S$ should be as large as possible. Since the elements of $S$ do not exceed 15, we first consider $\{15,14,13,12,11\}$. However, $15+12=13+14$, so if $15,14,13 \in S$, then $12 \notin S$. Removing 12 and replacing it with 11, we then examine whether 10 can be included in $S$. Since $15+10=14+11$, $10 \notin S$; similarly, $15+9=13+11$, so $9 \notin S$. Next, we consider 8, and after verification, we find that 8 can be included. Thus, $S=\{15,14,13,11,8\}$ meets the requirements, and the sum $15+14+13+11+8=61$ is the maximum value.