Minimal number of edges for complete linear hypergraphs A complete linear hypergraph is a hypergraph $H=(V,E)$ such that 


*

*$|e|\geq 2$ for all $e\in E$,

*$|e_1\cap e_2|=1$ for all $e_1, e_2\in E$ with $e_1\neq e_2$, and

*for all $v\in V$ we have $|\{e\in E:v\in e\}| \geq 2.$


For $n>2$ set $\mathbb{N}_n =\{1,\ldots,n\}$ and $$\ell(n)=\min\{|E|: E\subseteq{\cal P}(\mathbb{N}_n) \text{ and }(\mathbb{N}_n, E) \text{ is complete linear}\},$$ so $\ell(n)$ is the greatest lower bound for the number of edges on a complete linear hypergraph on $n$ points.
Question: What is the value of $\ell(n)$, depending on $n$?

(Note concerning least upper bounds: If $H=(V,E)$ is a complete linear hypergraph with $|V|=n>2$, then $|E| \leq n$ by the theorem of DeBruijn-Erdos, and we can reach $|E| = n$ with the so-called "near pencil", see the same link.)
 A: Let $L$ be the number of edges in the configuration and $n$ the number of points.  Then $n \leq {L \choose 2}$ and this bound is tight.
There are configurations with that many points:  Let the points correspond to the two element subsets of $\{1, 2, \ldots, k\}$.  And let the edges correspond to the numbers $\{1, 2, \ldots , k\}$.  Then say edge $L_i$ contains the vertex $v_{\{a,b\}}$ iff $i \in \{a,b\}$.  This is linear, and it has ${k \choose 2}$ points and $k$ edges.  It covers each point exactly twice.
Any configuration has at most that many points: The previous answer can be pushed to give $L \geq \sqrt{2n}$, but we can get our above bound exactly.  Let $\mathcal{L}$ be the set of edges.  For each point $i \in \{1, 2, \ldots , n\}$, associate a set $V_i \subseteq \mathcal{L}$, which is the edges in $\mathcal{L}$ containing the point $i$.  Then for all $i \neq j$, we have $|V_i \cap V_j| \leq 1$ (else there are two edges containing the same two points).  We know that each point is covered at least twice, so $|V_i| \geq 2$.  For each set $V_i$, let $W_i$ be any subset of size $2$ (it doesn't matter how we pick $W_i$).  Then the sets $W_i$ are distinct $2$-element subsets of $\mathcal{L}$, which means there are at most ${L \choose 2}$ of them.

Added afterwards: To complete the discussion, we have the corresponding lower bound on $n$ that $L \leq n$.  This is by (generalized) Fisher's inequality, and it uses the fact that the intersection of any two edges has size $1$.  And this can be attained by the near pencil mentioned in original post or a projective plane.
So together, we have $L \leq n \leq {L \choose 2}$ and both bounds are best possible.
A: It is about $\sqrt n$.
If each edge has size at most $\sqrt n$, then you need at least $2n/\sqrt n=2\sqrt n$ edges to cover everything twice.
If there's an edge of size at least $\sqrt n$, then you need at least these many other edges to cover each of its points at least twice.
With some more involved argument of this kind, you might even get asymptotically $2\sqrt n$.
