Solution:

We call a graph, satisfying the given condition, $n$-purple. Let $f(n)$ be the smallest possible number of blue edges in an $n$-purple graph.

Suppose that $f(n) < n + 5$ for $n \geq 5$. If any vertex of an $n$-purple graph $G$ with $f(n)$ blue edges is a head of at least two blue edges, then the total number of the blue edges is at least $2n$. Since $2n \geq n + 5 > f(n)$ for $n \geq 5$, we may find a vertex $a$ of $G$ that is the head of at most one blue edge. If $a$ is not head of a blue edge, then there is a vertex $b$ of the graph $G \setminus \{a\}$ that is a head of a blue edge and hence $G \setminus \{a, b\}$ is an $(n-1)$-purple graph. If $a$ and $b$ are joined by a blue edge, then $G \setminus \{a, b\}$ is an $(n-1)$-purple graph. In both cases the obtained $(n-1)$-purple graph has at least one blue edge less than $G$ and so $f(n) \geq f(n-1) + 1$. Then $f(n-1) < n + 4$, in particular, $f(4) < 9$.

Now we shall compute $f(4)$. It is well-known that there are $3$-purple graphs. Then the above arguments show that any vertex of a $4$-purple graph is a head of at least two blue edges. Hence $f(4) \geq 8$. If $f(4) = 8$, then any vertex is a head of exactly two blue edges. There are two such graphs containing no blue triangles but these graphs are not purple. If $f(4) = 9$, the two vertices are heads of three blue edges, and the remaining six vertices are heads of two blue edges. There are six such graphs containing no blue triangles but they are not purple. The following example shows that $f(4) = 10$ — a regular octagon with blue sides and two blue adjacent main diagonals, and the remaining diagonals are red.

So $f(n) \geq n + 5$ for $n \geq 5$. It is easy to see that the graph with $2n$ vertices such that its blue edges form three disjoint cycles with lengths $5, 5$ and $2n - 10$ is $n$-purple. Hence $f(n) = n + 5$ for $n \geq 5$.