Optimise the size of union of sets
Your problem is NP-hard. Given an instance $T_1,\ldots,T_n$ of Maximum Coverage, let $x_1,\ldots,x_n$ be new variables, and create an instance $S_1,\ldots,S_n$ of your problem, where $S_i = T_i \cup \{x_i\}$. We have
$$
\left| \bigcup_{i \in I} S_i \right| -|I| = \left| \bigcup_{i \in I} T_i \right|.
$$
On the positive side, consider the following greedy algorithm: $I_0 = \emptyset$, and for $k = 1,\ldots,n$, $I_k$ is obtained from $I_{k-1}$ by adding the set which maximizes the total number of elements covered by sets in $I_k$. It is known that for $k \in \{0,\ldots,n\}$, the set $I_k$ found by the greedy algorithm has the following property: for all sets $J_k$ of size $k$ we have
$$
\left| \bigcup_{i \in I_k} S_i \right|  \geq
(1-1/e) \left| \bigcup_{i \in J_k} S_i \right|.
$$
In particular, if the optimal solution to your problem has size $k$ and objective value $O$, then
$$
\left| \bigcup_{i \in I_k} S_i \right| \geq (1-1/e)(O + k),
$$
and so
$$
\left| \bigcup_{i \in I_k} S_i \right| - k \geq (1-1/e) O - k/e. 
$$
The minimal optimal $k$ satisfies $O \geq k$ (each set adds at least one new element), and so $(1-1/e) O - k/e \geq (1-2/e) O$. This shows that the greedy algorithm (more explicitly, finding $k$ which maximizes $|I_k|-k$) gives a $1-2/e$ approximation to your problem (this can probably be improved).