Call a graph satisfying the given condition $n$-magenta, and let $f(n)$ be the smallest number of edges in an $n$-magenta graph. Lemma. $f(n) \geq n+5, ~ \forall n \geq 5$. Proof. For $n \geq 5$, suppose for contradiction that $f(n) < n-5$. If all vertices in an $n$-magenta graph $G$ with$f(n)$ edges are incident to at least two blue edges, then the total number of blue edges is at least $2n \geq n+5 > f(n) ~ \forall n \geq 5$, so there is a vertex $v$ incident to at most one blue edge. If $v$ is not incident to a blue edge, then we can find a vertex $v'$ which is incident to a blue edge, and thus $G \setminus \{v,v' \}$ is $(n-1)$-purple. If $\overline{vv'}$ is blue, then $G \setminus \{v,v' \}$ is still $(n-1)$-purple. In either case, the obtained graph $G \setminus \{v,v'\}$ has at least one less blue edge than than $G$, so we get $f(n-1) < n+4$; in particular,$f(4)<9$. However, simple case-work shows that $f(4)=10$ with the following graph: [asy][asy] unitsize(3cm); for(int i=0;i<=7;i=i+1) { draw(dir(45*i)--dir(45*(i+1)),blue); draw(dir(45*i)--dir(45*(i+2))^^dir(45*i)--dir(45*(i+3))^^dir(45*i)--dir(45*(i+4)),red); draw(dir(45)--dir(225)^^dir(90)--dir(-90),blue); } for(int i=0;i<=7;i=i+1) { dot(dir(45*i)); } [/asy][/asy] This gives the required contradiction. $\Box$ Observe now that for $n \geq 5$, a graph with $2n$ vertices such that its $n+5$ blue edges form $n-3$ disjoint cycles of lengths $5,5,\underbrace{1,1, \dots ,1}_{\text{n-5}~1'\text{s}}$ is $n$-magenta, so it follows that $$\boxed{f(n)= \left \{ \begin{array}{l} 10~(\text{for}~n=4); \\ n+5~(\text{for}~n \geq 5). \end{array} \right.}$$

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