$A \subset S$ and all subsets of $A$ have distinct capacities, find the maximum capacity of $A$.

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If $|A| \geqslant 7$, then the number of its subsets with at most 4 elements is at least $\mathrm{C}_{7}^{1}+\mathrm{C}_{7}^{2}+\mathrm{C}_{7}^{3}+\mathrm{C}_{7}^{4}=98$, while $25+24+23+22=94$, so there must be two subsets with the same capacity. Contradiction! Therefore, $|A| \leqslant 6$.
When $|A| \leqslant 5$, its capacity does not exceed $25+24+\cdots+21=115$.
It is verified that: $\{25,24,23,21,18,12\}$ satisfies the condition and has a capacity of 123. If the capacity is greater than 123, then $|A|=6$. The sum of the two largest numbers is no more than 72, so the sum of the three smallest numbers is at least $124-72=52$. Since there are $\mathrm{C}_{6}^{3}=20$ three-element subsets, the three largest numbers must be $25,24,23$ (otherwise, the sum of the largest two numbers is less than 72, and the sum of the smallest three numbers is greater than 52, which means there must be two three-element subsets with the same capacity, contradiction!) and the sum of the smallest three numbers is 52 or 53. Let these three numbers be $b_{1}>b_{2}>b_{3}$, so $b_{1} \leqslant 21, b_{2} \leqslant b_{1}-3, b_{3} \leqslant b_{2}-3$. If $b_{1}+b_{2}+b_{3}=53$, it can only be $b_{1}=21, b_{2}=18, b_{3}=14$, but $25+14=21+18$, contradiction! If $b_{1}+b_{2}+b_{3}=52$, it can only be $b_{1}=21, b_{2}=18, b_{3}=13$ or $b_{1}=21, b_{2}=17, b_{3}=14$, but $21+18+73=25+24+13,24+14=21+17$, contradiction!
Therefore, the maximum capacity is 123.