Fixed size set to contain the maximum number of given sets
The decision version of the problem can be formulated as follows:


Given a collection of sets of numbers $S$. Each number is drawn from a set $U$.
You want to know if there is a way to pick $k$ numbers from $U$ to form a set $X$ which is the superset of $s$ sets in $S$.


The following is a reduction from k-clique which is an NP-complete problem.

The k-clique problem has an input as a graph and asks if there is a way to pick $k$ nodes so that there is an edge between any two of the $k$ nodes.

We turn each node into a number. These numbers form the set $U$. We turn each edge into a set containing two numbers representing the corresponding two nodes of the edge. These sets form the set $S$.

The question to the original problem is: Is there a way to pick $k$ numbers from $U$ to form a set $X$ such that $X$ is the superset of $C_{2}^k$ sets in $S$

The proof has two directions.


If there is a k-clique in the graph, then we can pick the $k$ numbers corresponding to $k$ nodes in the clique to form $X$. Because of the way we transform the graph, we know there will be $C_{2}^k$ sets in $S$ each corresponds to an edge in the original graph and thus is a subset of $X$. So the answer to the original problem is positive.
If there is no clique of size $k$, then for any set of $k$ nodes chosen from the graph one can only have less than $C_{2}^k$ undirected edges between any two picked nodes. Otherwise, we will have a clique of size $k$. This means that there can only be less than $C_{2}^k$ subsets of $S$ which is a subset of the set $X$ formed by any $k$ chosen numbers. So the answer to the original problem is not positive.


The transformation is in poly-time. The reduction is proven, so the problem is NP-hard.
The problem is clearly in NP because we check how many sets covered by $k$ numbers in poly time. This means the original problem is NP-complete.