We prove the following statement by induction on $n$:
If the vertices of a convex $4 n$-gon $P$ are colored red and blue such that each color occurs at least $(n-1)$ times, then we can find $(n-1)$ pairwise disjoint quadrilaterals $Q_{1}, \ldots, Q_{n-1}$, whose vertices are also vertices of $P$.
For $n=25$ we obtain the claim of the problem statement.

For $n=1$ there is nothing to show. Now let $n \geqslant 2$ and, without loss of generality, suppose there are at least $2 n$ red vertices. The case in which the vertices are colored alternately or 2-alternately is treated below. Otherwise, there must be four consecutive vertices among which there are not exactly two red and two blue. Then there also exist four consecutive vertices among which there are more red than blue, since otherwise there would be fewer red than blue vertices. Choose such four vertices: If exactly three of them are red, we choose the quadrilateral spanned by the four selected vertices as $Q_{n-1}$. The remaining $4(n-1)$ points form a convex $4(n-1)$-gon $P^{\prime}$, which is disjoint from $Q_{n-1}$ and to which we can apply the induction hypothesis. If, on the other hand, all four of the selected vertices are red, we proceed along the boundary of the $4 n$-gon until we hit for the first time one of the at least $n-1>0$ blue points, and apply the same argument as in the previous case.

There remain the cases in which the vertices are colored alternately or 2-alternately. For these there are various constructions. We first consider the alternating case and denote the vertices of $P$ by $P_{1}, P_{2}, \ldots, P_{4 n}$, where, without loss of generality, $P_{1}$ is blue. In this case there are (at least) two constructions (see the following sketches).

Variant 1

Variant 2

In Variant 1 we begin with the quadrilateral $P_{2} P_{3} P_{4} P_{4 n}$ and then continue from $P_{4}$ counterclockwise in steps of three and from $P_{4 n}$ clockwise in steps of one. In doing so, exactly ...