jmc

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 $$\begin{array}{rl}\frac{\frac{n(n+1)}{2}+2015}{n+2}& =\frac{n^2+n+4032}{2(n+2)}\\ & =\frac{1}{2}\left(n-1+\frac{4032}{n+2}\right)\\ & =\frac{1}{2}\left(n+2+\frac{4032}{n+2}\right)-\frac{3}{2}.\end{array}$$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 $\sqrt{4032}-2\approx \mathrm{61.5.}$

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