The answer is $\lceil\frac nk\rceil$. Let $S=\prod_{i=1}^m p_i^{\alpha_i}$ and the $k$-th largest divisor of $S$ be $\prod_{i=1}^m p_i^{\beta_i}$. Note that every number of the form $\prod_{i=1}^m p_i^{\gamma_i}$, where $\gamma_i\in[\beta_i,\alpha_i]$ must be among the $k$ largest divisors of $S$, so $k\ge\prod_{i=1}^m(\alpha_i-\beta_i+1)$. Now we use the somewhat trivial inequality $(b+1)(a-b+1)\ge a+1(\iff b(a-b)\ge 0)$ to get that $$k\prod_{i=1}^m(\beta_i+1)\ge \prod_{i=1}^m(\alpha_i-\beta_i+1)\prod_{i=1}^m(\beta_i+1)=\prod_{i=1}^m (\alpha_i-\beta_i+1)(\beta_i+1)\ge\prod_{i=1}^m(\alpha_i+1)=n$$so the desired number of divisors, $\prod_{i=1}^m(\beta_i+1)$, is at least $\lceil\frac nk\rceil$. For the construction, take $S=p^{k-1}q^{\lceil\frac nk\rceil-1}$, where $q>p^{k-1}$. Then the $k$-th largest divisor of $S$ is $q^{\lceil\frac nk\rceil-1}$.