Question:
Let $S$ be a set containing distinct integers such that the smallest element is 0 and the largest element is 2015.  Find the smallest possible average of the elements in $S.$

Answer:
It is clear that to get the smallest positive average, the set should be of the form $S = \{0, 1, 2, \dots, n, 2015\}$ for some nonnegative integer $n.$  For this set, the average is
\begin{align*}
\frac{\frac{n(n + 1)}{2} + 2015}{n + 2} &= \frac{n^2 + n + 4032}{2(n + 2)} \\
&= \frac{1}{2} \left( n - 1 + \frac{4032}{n + 2} \right) \\
&= \frac{1}{2} \left( n + 2 + \frac{4032}{n + 2} \right) - \frac{3}{2}.
\end{align*}By AM-GM,
\[\frac{4032}{n + 2} + n + 2 \ge 2 \sqrt{4032}.\]However, equality cannot occur, since $n + 2 = \sqrt{4032}$ does not lead to an integer, so we look for integers close to $\sqrt{4032} - 2 \approx 61.5.$

For both $n = 61$ and $n = 62,$ the average works out to $\boxed{62},$ so this is the smallest possible average.