Find the minimum number of edges in a graph with $3n+1$ vertices if ... 
Let $G$ be a simple graph with $3n+1$ vertices. For any vertex $v$, there exists $n$ disjoint $K_3$ (i.e. triangle) such that none of them contains $v$. Find minimum number of edges of graph $G$.


*

*If we take $n$ triangles $v_iu_iw_i$, where $1\leq i\leq n$ and a vertex $v_0$ connected with each vertex we get a graph which satysfies the condition so $\varepsilon _{\min} \leq 6n$ .


*Now suppose there is a graph with $6n-1$ edges or less.
Lemma Let $v$ be a vertex with minimum degree
$d$ in $G$. Then $d=3$.
Proof: By handshake lemma we have $3$: $$(3n+1)d \leq \sum _{i=1}^{3n+1} d_i \leq 12n-2\implies d \leq {12n-2\over 3n+1}<4$$
On the other hand if $d <3$ then $v$ has at most $2$ neighbours $u$ and $w$, so for $u$ we have a triangle which does not contain $u$ and does contain $v$. But then $v$ has another neighbour in this triangle (beside $w$). So $v$ has exactly $3$ neighbours.
Claim: Every edge is present in some triangle.
If not then removing it we get a smaller graph with
the property same as $G$ has.
Conjectures: Minimal configuration has exactly
$4n$ triangles.
Last edit before the end of bounty.
Any suggestion how to continue?
 A: A little bit ahead of me.
But that's the way it's going to be.
Lemma 1.
Let $G$ be a graph with this property.
If $a$ is a vertex of degree 3 and $b,c,d$, its neighbors,
then $a,b,c,d$ form graph $K_4$.
Proof.
If we remove vertex $d$, then the only triangle containing vertex $a$ is $a,b,c$.
So there is an edge $bc$.
Lemma 2.
Let $a$ be a vertex of degree 3 and $b,c,d$, its neighbors.
Let $G'=G/\{a,b,c,d\}$ be obtained from $G$ by contracting vertices $a,b,c,d\ $ to vertex $a'$
(we remove vertices $a,b,c,d$ and add a new vertex $a'$,
where the edges incident to $a'$ each correspond to an edge incident to
either $b$, or $c$, or $d$).
Then $G'$ is a graph with our property.
Proof.
If we remove vertex $a'$ from graph $G'$,
then the desired triangles are all those triangles
which are obtained by removal of vertex $d$ in the original graph $G$
except for triangle $a,b,c$.
Almost similarly we find triangles if we remove vertex $x\neq a'$.
A: You already have several good ideas, but you are missing point 5 below.
Let's given a name to the property:
$(P)$ For any vertex $v$, there exists $n$ disjoint $K_3$ (i.e. triangle) such that none of them contains $v$.

*

*Find a family of graphs $G_n$ on $3n+1$ vertices with property $(P)$, such that $G_n$ has $6n$ edges for every $n \geq 0$
Let $G = (V,E)$ be a graph with property $(P)$ on $3n+1$ vertices and with a minimum number of edges.


*Using question 1, prove that there must be a vertex of degree at most 3.


*Prove that every vertex has degree at least 3.


*Let $x$ be vertex of degree 3. Prove that the set of neighbors $N(x)$ of $x$ induces a triangle $uvw$.


*Let $H = (V_H,E_H)$ be the graph obtained from $G$ by contracting $x,u,v,w$ into a single vertex. Prove that $H$ has property $(P)$. To be clear,
$$V_H := V \setminus \{u,v,w\}\quad\textrm{and}$$
$$E_H := \{ yz~:~y,z \in V_H \setminus \{x\}, yz \in E\} \cup \{ yx~:~ y \in V_H \setminus \{x\}, \{yu,yv,yw\} \cup E \neq \emptyset \}$$


*Find the relationships between $|E_H|$ and $|E|$ and between $|V_H|$ and $|V|$. By induction on $n$, prove that $G$ has $6n$ edges.
