Where can I get 30 other people?

Let $a_n$ = the number of subsets with $k$ elements such that $S \equiv n \pmod{31}$ . Also, let $z = e^{2 \pi i/31}$ . We have

$\sum_{n = 0}^{30} a_n z^n = \sum_{1 \leq d_1 < d_2 < \cdots < d_k \leq 2015}^{} z^{d_1 + d_2 + \cdots + d_k},$

where the $d_i$ s are the elements of the subset. However, this is also the coefficient of $x^{2015 - k}$ of

$(x + z)(x + z^2)\cdots (x + z^{2015}).$

The above expression can be simplified using $z^{31} = 1$ to get

$\begin{aligned} (x + z)(x + z^2) \cdots (x + z^{2015}) &= [(x + 1)(x + z)\cdots (x + z^{30})]^{65} \\ &= (x^{31} + 1)^{65}, \end{aligned}$

which, from binomial expansion, becomes

$x^{2015} + \binom{65}{1} x^{1984} + \cdots + \binom{65}{64} x^{31} + 1.$

For $k = 0, 31, 62, \dots, 2015$ , we get that $\sum_{n = 0}^{30} a_n z^n = c$ , where $c > 0$ . We claim that $a_0 - c = a_1 = a_2 = \cdots = a_{30}$ , so $a_0$ is greater than the other $a_i$ s. Since all subsets are chosen with equal probability, the person with the integer $0$ has a greater chance of winning than the others. Thus, the sum of the possible $k$ is

$0 + 31 + 62 + \cdots + 2015 = 31(1 + 2 + \cdots + 65) = 31(33)(65) = \boxed{66495}.$