Using the prime correspondence in localizations which has already been mentioned, and Prime avoidance, we can show that the maximal ideals are exactly the $S^{-1}P_i$.
Suppose $J$ is any ideal strictly containing a $P_j$. We claim that $J\cap S\neq \emptyset$. If it were otherwise, then $J\subseteq \cup_i P_i$, and by prime avoidance, $J$ must be contained in a $P_j$. Obviously $J$ is not contained in $P_i$, so $J\subseteq \cup_{j\neq i}P_j$. But then $P_i\subset J\subseteq \cup_{j\neq i}P_j$, a contradiction. The upshot of this is that the union hypothesis forces ideals strictly containing any $P_j$ to hit $S$ and therefore explode in the localization.
You now apply this to show that each $S^{-1}P_i$ is maximal in $S^{-1}R$. If $P_i$ were not maximal, then a maximal ideal sitting above $P_i$ would contract to a prime ideal strictly containing $P_i$, but such an ideal can't be proper, as we've seen.
You now have at least $n$ maximal ideals of $S^{-1}R$. All that remains is to show there are no more.
If $M$ is maximal in $S^{-1}R$, it contracts to a prime $Q$ in $R$ such that $Q\cap S=\emptyset$. Thus $Q\subseteq \cup P_i$ and by prime avoidance $Q\subseteq P_i$ for some $i$. It follows that $S^{-1}Q=M=S^{-1}P_i$ by maximality of $M$, and we have indeed only $n$ maximal ideals.