Solution:

We have to prove that if the edges of a complete graph with 9 vertices are colored in blue and red in such a way that there is no blue quadrilateral then its vertices can be partitioned into 4 groups without any blue edges inside any group.

Lemma 1. The edges of a complete graph with 6 vertices are colored in blue and red in such a way that there is no blue quadrilateral and its vertices can not be partitioned into 3 groups without blue edges inside any group. Then the graph does not contain a red triangle.

Proof. It is well known that the complete graph with 6 vertices and edges colored in two colors (red and blue) has an one-colored triangle. Denote the vertices of the graph by $v_{1}, v_{2}, \ldots, v_{6}$ and assume the existence of a red triangle $v_{1} v_{2} v_{3}$. If some edge $v_{i} v_{j}, i, j \in\{4,5,6\}$, is red, then $\{v_{1}, v_{2}, v_{3}\},\{v_{i}, v_{j}\}$ and $\{v_{k}\}, k \neq 1,2,3, i, j$, is a partition which is supposed not to exist.
Therefore $v_{4} v_{5} v_{6}$ is a blue triangle. Now the condition implies that for every $i=1,2,3$ at least one of the edges $v_{i} v_{j}, j=4,5,6$, is red. If the edges $v_{1} v_{4}$ and $v_{2} v_{4}$ are red then $v_{1} v_{2} v_{4}$ is a red triangle and, as above, $v_{3} v_{5} v_{6}$ is a blue triangle. Now, if the edge $v_{3} v_{4}$ is blue, then $v_{3} v_{4} v_{5} v_{6}$ is a blue quadrilateral and, if the edge $v_{3} v_{4}$ is red, then we have the partition $\{v_{1}, v_{2}, v_{3}, v_{4}\},\{v_{5}\}$ and $\{v_{6}\}$, a contradiction, which completes the proof of the lemma.

Lemma 2. Under the conditions of Lemma 1 all red edges of $G$ form a cycle of length 5.

Proof. We know from Lemma 1 that $G$ does not contain a red triangle.
Without loss of generality we can assume that the edges $v_{1} v_{2}$ and $v_{3} v_{4}$ are red. If all edges $v_{i} v_{j}$ for $i=1,2, j=3,4$ are red, then the partition $\{v_{1}, v_{2}, v_{3}, v_{4}\},\{v_{5}\}$ and $\{v_{6}\}$ ensures a contradiction. So, let $v_{2} v_{4}$ be blue. Since the quadrilateral $v_{2} v_{4} v_{5} v_{6}$ can not be blue, we may assume without loss of generality that the edge $v_{4} v_{6}$ is red. Now $v_{3} v_{6}$ is blue (otherwise $v_{3} v_{4} v_{6}$ is a red triangle) and $v_{3} v_{5}$ is blue (otherwise we have the partition $\{v_{1}, v_{2}\},\{v_{3}, v_{5}\}$ and $\{v_{4}, v_{6}\}$ for a contradiction).
If the edge $v_{1} v_{3}$ is red then using the conditions of the problem we see that the edge $v_{2} v_{5}$ is blue, the edge $v_{2} v_{6}$ is red and the edges $v_{1} v_{6}, v_{1} v_{5}, v_{4} v_{5}$ and $v_{1} v_{4}$ are blue. We finally conclude that the red edges in $G$ are $v_{1} v_{2}, v_{2} v_{6}, v_{6} v_{4}$, $v_{4} v_{3}$ and $v_{3} v_{1}$ and they form a cycle of length 5 as desired.
If the edge $v_{1} v_{3}$ is blue, then similar arguments lead to a contradiction. This completes the proof of the lemma.

Let us now consider a graph with 9 vertices $v_{1}, v_{2}, \ldots, v_{9}$, satisfying the given conditions. It is known that such a graph contains a red triangle or a blue quadrilateral. Since the latter is impossible we have a red triangle $v_{7} v_{8} v_{9}$, say. If the induced graph $v_{1} \ldots v_{6}$ can be divided into three groups without blue edges inside any group then we obtain the required division of $G$.
Otherwise Lemma 2 implies that we may assume without loss of generality that the edges $v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}, v_{4} v_{5}$ and $v_{5} v_{1}$ are red.
If the edges $v_{i} v_{6}, i=7,8,9$, are red then $\{v_{6}, v_{7}, v_{8}, v_{9}\},\{v_{1}, v_{3}\},\{v_{2}, v_{4}\}$ and $\{v_{5}\}$ is the desired division. If the edge $v_{7} v_{6}$ is blue while the edges $v_{8} v_{6}$ and $v_{9} v_{6}$ are red then at least three of the edges $v_{7} v_{i}, i=1,2, \ldots, 5$, are red and we easily have the desired division (otherwise we get a red quadrilateral).
Similar arguments solve the two remaining cases: when two of the edges $v_{6} v_{7}, v_{6} v_{8}$ and $v_{6} v_{9}$ are blue and one is red, and when all of them are blue.