A parital solution:
Let $n \in \mathbb{N}$ and let $\mathcal{F}$ such that $\forall i,j \in [n]\exists S\in \mathcal{F}:i,j\in S$ and that $S$ is unique.
Denote by $t$ the size of the biggest set in $\mathcal{F}$ and assume WLOG that the set is $S=\{1,...,t\}$.
Case 1: $t\ge \frac{n}{2}$:
For each $i \in S$, we must have a set containing the pair $(t+1,i)$, and as we can't have $2$ elements from $S$ together in another set, we must have $t$ new sets.
As $t+2$ can be in at most $1$ of the above sets, we need at least $t-1$ new sets for $t+2$ (to have the edges $(t+2,i): i\in S$). We can continue similarly until $n=t+(n-t)$ intoduces $t-(n-t-1)$ new sets.
We got a total of $$1+t(n-t)-(0+1+...+n-t-1)=1+t(n-t)-\frac{(n-t)(n-t-1)}{2}=$$ $$ 1+\frac{n-t}{2}(3t-n+1)$$ sets.
Treating $t$ as a variable and $n$ fixed, we can see that the function has a global maxima at $t \sim \frac{2}{3}n$ so it is enough to check for $t=n-1, \frac{n}{2}$.
For $t=n-1$ we get $n$ subsets, which is exactly enough.
For $t=\frac{n}{2}$ we get $\ge \frac{n}{4}(\frac{n}{2}+1)$. We want it to be $\ge n$ so we need $\frac{n}{2} + 1 \ge 4$ or $n\ge 6$.
So we solved for this case for every $n$ above $5$, and for $n=2,3,4,5$ it can be verified manually.
Case 2: $t < \frac{n}{2}$
First we can observe that if there are less than $n$ subsets in $\mathcal{F}$, then $t\ge \sqrt n$:
Each set contributes at most $t \choose 2$ pairs and we have at most $n-1$ subsets, and so at most $(n-1)\binom{t}{2}$. On the other hand, we know there are exactly $n \choose 2$ pairs. So, we got that $$\binom{n}{2}\le (n-1)\binom{t}{2}\rightarrow n\le t(t-1)\le t^2 \rightarrow t\ge \sqrt n$$
We can use a similar method to the previous case to deduce that there are at least $1+2+...+t$ other subsets, so there are at least $\binom{\sqrt{n} + 1}{2}\ge \frac{n}{2}$ subsets.
I was not able to improve the bound in this case.
Hope it gives you an idea that can generalize it to a complete solution.