jmc

We find the sum of all possible hundreds digits, then tens digits, then units digits. Every one of $\{1,2,3,4,5,6,7,8,9\}$ may appear as the hundreds digit, and there are $9\cdot 8=72$ choices for the tens and units digits. Thus the sum of the hundreds places is $(1+2+3+\cdots +9)(72)\times 100=45\cdot 72\cdot 100=324000$. Every one of $\{0,1,2,3,4,5,6,7,8,9\}$ may appear as the tens digit; however, since $0$ does not contribute to this sum, we can ignore it. Then there are $8$ choices left for the hundreds digit, and $8$ choices afterwards for the units digit (since the units digit may also be $0$). Thus, the the sum of the tens digit gives $45\cdot 64\cdot 10=28800$. The same argument applies to the units digit, and the sum of them is $45\cdot 64\cdot 1=2880$. Then $S=324000+28800+2880=355\boxed{680}$.