Let $A=\{1,2,...,n\}$. Let the powerset of $A$ be $\mathcal{P}(A)$.
We call $S\subseteq \mathcal{P}(A)$ a paired family of subsets if $\forall a\in A$, the number of elements of $S$ that contain $a$ is even.
For $|A|=n$, how many $S\subseteq \mathcal{P}(A)$ are paired?
Answer is $2^{2^n-n}$.
I'll give two (morally the same) proofs.
Proof Using Linear Algebra
Identifying subsets with indicator functions, we can view every $\mathcal{S}\subseteq\mathcal{P}(A)$ as an element of the $\mathbb{F}_2$-vectorspace $(\mathbb{F}_2)^{\mathcal{P}(A)}$. For each $a\in A$, define an $\mathbb{F}_2$-linear map $p_a\colon (\mathbb{F}_2)^{\mathcal{P}(A)}\to\mathbb{F}_2$ by $$S\in\mathcal{P}(A)\mapsto \begin{cases}1 & a\in S\\0 & a\notin S\end{cases}$$ so $p_a$ is counting whether $a$ appears in even or odd number of elements $S\in\mathcal{S}\subseteq\mathcal{P}(A)$ (exactly as in the condition of paired families). So counting paired families becomes a problem of counting elements in the common kernel of all $p_a$'s.
The set $\{p_a\mid a\in A\}$ is linearly independent (easy check), hence there are $n$ linearly independent conditions being put on $(\mathbb{F}_2)^{\mathcal{P}(A)}$. So the common kernel has dimension $\lvert\mathcal{P}(A)\rvert-n=2^n-n$ over $\mathbb{F}_2$, so $2^{2^n-n}$ paired families.
Proof Using Simple Counting/Probability
Without loss of generality, let $A=\{1,2,3,\dots,n\}$. Suppose we ask the question: What proportion of families will contain $1$ even number of times? We can argue the answer is one-half by the following observation: Pair off the elements of $\mathcal{PP}(A)$ as $\mathcal{S},\mathcal{S}\triangle\{\{1\}\}$, i.e. the pairs agree on all subsets of $A$ except the singleton $\{1\}$. Exactly one member from each pair will contain $1$ even number of times.
Continuing this argument for $2$, exactly one from each $$\mathcal{S},\mathcal{S}\triangle\{\{2\}\}$$ will have even number of $2$'s. Moreover, this is independent of the the number of $1$s, so a quarter of families have both even number of $1$'s and $2$'s.
Repeat for $3,\dots,n$ therefore gives us the probability of each element $1,2,\dots,n$ appearing even number of times is $2^{-n}$, because we have $\{1\},\{2\},\dots,\{n\}$ all can appear (or not) independently.
So the total number of paired families is $2^{-n}\cdot\lvert\mathcal{PP}(A)\rvert=2^{2^n-n}$.