, 2016

Nora has a winning strategy.
Let Mable write down $p_1 < p_2 < \dots < p_{2015}$ in step (1). Then $P = p_1p_2\dots p_{2015}$ is Mable's score.
Let $q_k = p_1p_2\dots p_k$ for each $k$. Nora needs to write down a positive integer $Q$ in step (2) such that $(P, Q + q_j) = 1$ for all $j$. Indeed, for each $1 \le k \le 2015$, we choose $r_k \in \mathbb{Z}$ such that
$$r_k \ne -q_j \pmod{p_k}$$
for all $1 \le j \le 2015$. Such an $r_k$ exists because there are at most $k < p_k$ residues that $r_k$ cannot take (as $p_k \mid q_j$ if $j \ge k$).

Now, there exists $Q \in \mathbb{Z}^+$ such that $Q \equiv r_k \pmod{p_k}$ for all $k$ by the Chinese remainder theorem. For this choice of $Q$, we have
$$Q + q_j \equiv r_k + q_j \not\equiv 0 \pmod{p_k}$$
for any $j$ and $k$. Therefore, we have $(P, Q + q_j) = 1$ for all $j$ as desired.