imc

To get from $S_n$ to $S_{n+1}$, we add $1(n+1)+2(n+1)+\cdots +n(n+1)=(1+2+\cdots +n)(n+1)=\frac{n(n+1{)}^2}{2}$. Now, we can look at the different values of $n$ mod $3$. For $n\equiv 0(\mathrm{mod}3)$ and $n\equiv 2(\mathrm{mod}3)$, then we have $\frac{n(n+1{)}^2}{2}\equiv 0(\mathrm{mod}3)$. However, for $n\equiv 1(\mathrm{mod}3)$, we have $$\frac{1\cdot 2^2}{2}\equiv 2(\mathrm{mod}3).$$ Clearly, $S_2\equiv 2(\mathrm{mod}3).$ Using the above result, we have $S_5\equiv 1(\mathrm{mod}3)$, and $S_8$, $S_9$, and $S_{10}$ are all divisible by $3$. After $3\cdot 3=9$, we have $S_{17}$, $S_{18}$, and $S_{19}$ all divisible by $3$, as well as $S_{26},S_{27},S_{28}$, and $S_{35}$. Thus, our answer is $8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{197}.$