Let $n$ be an integer greater than $1$ and let $S$ be a finite set containing more than $n+1$ elements.Consider the collection of all sets $A$ of subsets of $S$ satisfying the following two conditions :
[b](a)[/b] Each member of $A$ contains at least $n$ elements of $S$.
[b](b)[/b] Each element of $S$ is contained in at least $n$ members of $A$.

Determine $\max_A \min_B |B|$ , as $B$ runs through all subsets of $A$ whose members cover $S$ , and $A$ runs through the above collection.
We need to determine the maximum value of $\min_B |B|$, where $B$ runs through all subsets of $A$ whose members cover $S$, and $A$ runs through the collection of subsets of $S$ satisfying the given conditions (a) and (b).

1. **Understanding the Problem:**
   - Let $S$ be a finite set with $|S| = m$ and $m > n + 1$.
   - Let $\mathcal{A}$ be a collection of subsets of $S$ such that:
     - Each member of $\mathcal{A}$ contains at least $n$ elements of $S$.
     - Each element of $S$ is contained in at least $n$ members of $\mathcal{A}$.

2. **Constructing a Cover:**
   - We need to find a subset $B \subseteq \mathcal{A}$ such that the union of the sets in $B$ covers $S$.
   - We aim to minimize $|B|$ and then maximize this minimum over all possible collections $\mathcal{A}$.

3. **Upper Bound:**
   - Consider any $A \in \mathcal{A}$. $A$ covers at least $n$ elements of $S$.
   - For the remaining $m - n$ elements of $S$, we need to find sets in $\mathcal{A}$ that cover these elements.
   - Since each element of $S$ is contained in at least $n$ members of $\mathcal{A}$, we can always find a set in $\mathcal{A}$ that covers each of the remaining $m - n$ elements.
   - Therefore, we can cover $S$ with at most $|S| - n + 1$ sets from $\mathcal{A}$.

4. **Lower Bound:**
   - We need to show that there exists a collection $\mathcal{A}$ such that no fewer than $|S| - n$ sets are required to cover $S$.
   - Consider $S = \{1, 2, \ldots, m\}$ and construct $\mathcal{A}$ as follows:
     \[
     \mathcal{A} = \left\{ A \subset S : A = (\{1, 2, \ldots, n\} \setminus \{x\}) \cup \{y\} \mid x \in \{1, 2, \ldots, n\}, y \in \{n+1, \ldots, m\} \right\}
     \]
   - Each set in $\mathcal{A}$ contains exactly $n$ elements.
   - Each element in $\{1, 2, \ldots, n\}$ is contained in $m - n$ sets, and each element in $\{n+1, \ldots, m\}$ is contained in $n$ sets.
   - To cover $S$, we need at least $m - n$ sets from $\mathcal{A}$.

5. **Conclusion:**
   - We have shown that $\max_A \min_B |B| \leq |S| - n$ and constructed an example where $\max_A \min_B |B| = |S| - n$.
   - Therefore, the required number is $|S| - n$.

The final answer is $\boxed{|S| - n}$.