Does a certain points and lines configuration exist? For which $n$ we may mark $n$ red and $n$ blue points on the Euclidean plane, not all on a line, so that any line which passes through two points of different colour contains another point?  
For $n=1991$ this was proposed in a not-up-to-date edition of Prasolov's problem book on planimetry, but the suggested solution actually solves a different problem (in the newest edition this is fixed.) 
The following example for $n=6k$ is communicated by M. Belozerov: take a regular $4k$-gon, colour its vertices alternatively and infinite points of the sides arbitrarily.
 A: OK. I'll start the list of answers with an example for n=4.  Look at a picture of a Fano plane.  It is an equilateral triangle where I will color the 3 vertices red, the 3 edge midpoints blue, and the triangle center also blue.  We need a fourth red point, so put it between two of the blue midpoints so that it is on the line between the third midpoint and the opposite red vertex.  So we have decorated the Fano plane by coloring its 7 points and adding an eighth point.
I initially thought of the pentagram (10 trees in five rows of four puzzle) but I run out of points; even when I add a central point I can't find a good place for a matching point.  So n=5 or 6 still escape me.
Gerhard "Let The Point Configurations Roll" Paseman, 2017.07.20.
A: Using a Pappian-ish configuration, we can get $n = 5$ and $n = 7$.
For $n = 7$, put points at $(x,0)$ and $(x,1)$ for integer $x$ with $0 \le x \le 3$, at $(x,1/2)$ for integer or half-integer $x$ with $1/2 \le x \le 5/2$, and at horizontal infinity. Color them like this:

For $n = 5$, remove the rightmost two blue points and the rightmost two red points, turn the rightmost blue point on the central horizontal line red, and make one of the red points on the middle vertical line blue. This construction breaks down for $n > 7$ since you start to get lines of shallow enough slope to just connect two points lying entirely in one "half" of the diagram.
Edited following the suggestion of Jan Kyncl:
A modification of the above in fact gives a configuration for any $n \ge 4$. For $n = 3k+1$ with $k \ge 1$, put points at $(x,0)$ and $(x,1)$ for integer $0 \le x \le 2k-1$, at $(x,1/2)$ for integer or half-integer $(k-1)/2 \le (3k-1)/2$, and at horizontal infinity. Then color the points on the horizontal lines $y = 0$ and $y = 1$ red if $x < k$ and blue if $x \ge k$, color the points on the horizontal line $y = 1/2$ blue if $x < k$ and red if $x \ge k$, and color horizontal infinity red. (This directly extends the $n = 7$ case drawn above)
For $n = 3k$ with $k \ge 2$, just omit the leftmost and rightmost point on the central horizontal line.
For $n = 3k + 2$ with $k \ge 1$, put points at $(x,0)$ and $(x,1)$ with integer $0 \le x \le 2k$, at $(x,1/2)$ with $k/2 \le x \le 3k/2$, and at horizontal infinity. On the outer lines $y = 0$ and $y = 1$, color points red if $x < k$ and blue if $x > k$, and on the central line $y = 1/2$ color points red if $x > k$ and blue if $x < k$. On the line $x = k$, color $(k,1)$ red, $(k,1/2)$ and $(k,0)$ blue, and color horizontal infinity red.
