Denote the vertices of the triangle by $A$, $B$ and $C$ and let $A'$, $B'$ and $C'$ be the midpoints of the sides $BC$, $CA$ and $AB$. At least two of the points $A'$, $B'$ and $C'$ are the same colour. We may assume that the points $A'$ and $B'$ are both red. The segment $A'B'$ is parallel to the segment $AB$. If we can find two red points on the segment $AB$ then we have found the trapezoid we wanted. Otherwise at most

one of the points on the segment is red. If there exists a red point distinct from $A$ and $B$, then denote it by $D$. If there is none, then let $D$ be an arbitrary point on the segment $AB$ distinct from $A$ and $B$. Hence, all the points on the segment $AB$ with the exception of $A$, $B$ and possibly $D$, are blue.

Assume that there exist two distinct blue points on the segment $AC$ also both distinct from $A$. Denote them by $E$ and $F$, so that $E$ is the one closer to $A$. Let $F'$ be a blue point on the segment $AD$ distinct from $A$. Let $E'$ be the intersection of the segment $AD$ and the line through $E$ parallel to the line $FF'$. Then $EE'F'F$ is a trapezoid and all of its vertices are blue.

Similarly, we show that either we can find such a trapezoid or else there is
at most one blue point other than $B$ on the segment $BC$. In the latter case the segments $AC$ and $BC$ are almost entirely red and we can easily find a trapezoid with the required property. We can get its vertices as the intersections of two parallel lines with the segments $AC$ and $BC$.