= Solution
Take $S$ to be the <primes> in the interval and $Q=\max(2,\lfloor\sqrt N\rfloor)$. Every such <prime> exceeds $M>N\ge Q$ apart from harmless bounded small cases, so it avoids zero modulo each <prime> up to $Q$. The sifted-interval bound with $H=N$ and $q=1$ gives
$$
\boxed{\pi(M+N)-\pi(M)\ll N/\log N.}
$$
Since $Q^2=O(N)$ and $\log(2Q)\asymp\log N$, the implied constant is absolute. The choice of strict or inclusive endpoint in the prime-counting convention changes at most two terms, absorbed by the bound for $N\ge2$.
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