Example 3 For a set of numbers $M$, define the sum of $M$ as the sum of all numbers in $M$, denoted as $S(M)$. Let $M$ be a set composed of several positive integers not greater than 15, and any two disjoint subsets of $M$ have different sums. Find the maximum value of $S(M)$.
Let $M=\{15,14,13,11,8\}$, at this point, $S(M)=61$. Below, we prove: for any set $M$ that meets the problem's conditions, $S(M) \leqslant 61$. For this, we first prove: $|M| \leqslant 5$. (1)

By contradiction, assume $|M| \geqslant 6$, and consider all subsets $A$ of $M$ with no more than 4 elements, then $S(A) \leqslant 15+14+13+12=54$, and the number of such subsets $A$ is $\mathrm{C}_{6}^{1}+\mathrm{C}_{6}^{2}+\mathrm{C}_{6}^{3}+\mathrm{C}_{6}^{4}=56$, thus there must be two subsets $A$ and $B$ such that $S(A)=S(B)$. Let $A^{\prime}=A \backslash(A \cap B)$, $B^{\prime}=B \backslash(A \cap B)$, then $A^{\prime}$ and $B^{\prime}$ are disjoint, and $S\left(A^{\prime}\right)=S\left(B^{\prime}\right)$, which contradicts the problem's conditions. Therefore, $|M| \leqslant 5$.

Consider any set $M$ that meets the conditions.
(1) If $15 \notin M$, then by (1), $S(M) \leqslant 14+13+12+11+10=60$;
(2) If $14 \notin M$, then by (1), $S(M) \leqslant 15+13+12+11+10=61$;
(3) If $13 \notin M$, note that $15+11=14+12$, so $M \neq\{15,14,12,11,10\}$, thus by (1), $S(M)<15+14+12+11+10=62$;
(4) If $15,14,13 \in M$, then $12 \notin M$.
(i) If $11 \in M$, then $10,9 \notin M$. Then by (1), $S(M) \leqslant 15+14+13+11+8=61$;
(ii) If $11 \notin M$. Then by (1), $S(M) \leqslant 15+14+13+10+9=61$.

In summary, $S(M) \leqslant 61$. Therefore, the maximum value of $S(M)$ is 61.