For JEE Advanced beginners

At first, we need to determine the last term of the sequence. $\begin{aligned}\sum n&=\frac{n(n+1)}{2}\\&=2015\\\Rightarrow n&=62.98\end{aligned}$ However, given is a series of integers. There are $r$ number of $r$ terms. Hence, the sequence will contain numbers from $1$ to $63$ with the number $63$ repeated $62$ times. Hence the series is $\begin{aligned}\sum_{i=1}^{63}{i}^2-63&=\frac{63(64)(127)}{6}-63\\&=85281\end{aligned}$ The sum of the digits of $S$ equals to $\Huge\color{#3D99F6}{\boxed {24}}$