Solution:
For each subset $A$ of $\{1, 2, \ldots, n-1\}$, we pair it with $A \cup \{n\}$. Note that for any such pair $(A, B)$ not both $A$ and $B$ can be in $\mathcal{F}$. Since there are $2^{n-1}$ such pairs it follows that $\mathcal{F}$ can have at most $2^{n-1}$ elements.

We shall show by induction on $n$ that if $|\mathcal{F}| = 2^{n-1}$ then $\mathcal{F}$ contains either all the subsets with odd number of elements or all the subsets with even number of elements. The result is easy to see for $n = 1$. Suppose that the result is true for $n = m-1$. We now consider the case $n = m$. Let $\mathcal{F}_1$ be the set of those elements in $\mathcal{F}$ which contain $m$ and $\mathcal{F}_2$ be the set of those elements which do not contain $m$. By induction, $\mathcal{F}_2$ can have at most $2^{m-2}$ elements. Further, for each element $A$ of $\mathcal{F}_1$ we consider $A \setminus \{m\}$. This new collection also satisfies the required property, so it follows that $\mathcal{F}_1$ has at most $2^{m-2}$ elements. Thus, if $|\mathcal{F}| = 2^{m-1}$ then it follows that $|\mathcal{F}_1| = |\mathcal{F}_2| = 2^{m-2}$. Further, by induction hypothesis, $\mathcal{F}_2$ contains all those subsets of $\{1, 2, \ldots, m-1\}$ with (say) even number of elements. It then follows that $\mathcal{F}_1$ contains all those subsets of $\{1, 2, \ldots, m\}$ which contain the element $m$ and which contain an even number of elements. This proves that $\mathcal{F}$ contains either all the subsets with odd number of elements or all the subsets with even number of elements.