**DEFINITIONS:**   Functions   $c : \binom X3\rightarrow \{0\ 1\}$   are called 2-colorings of triangles in   $X$.   The $4$-element subsets   $A\subseteq X$   are called tetrahedra. Each 2-coloring   $c$   of triangles induces   $(\alpha\ \beta)$-coloring of each tetrahedron   $A$,   where
$$\beta := \sum\_{T\subseteq A,\ |T|=3}\ c(T)\quad\quad\quad \alpha := 4-\beta$$


**QUESTION:**   Does every set   $X$   admit a 2-coloring   $c$   of its triangles such that the induced coloring of every tetrahedron is of the   $(2\ 2)$ type?   And if the answer is NOT, then what is the smallest cardinality   $|X|$   for which   $X$   does not admit such   $c$   (then such cardinality must be finite)?


**A PARTIAL RESULT:**   If   $|X| \le 6$   then there exists a 2-coloring of triangles such that all tetrahedra are colored   $(2\ 2)$.


(Of course, if such a coloring exists for a set $X$ then it exists--it induces--a similar coloring of triangles in every subset of $X$, hence together with any *good* cardinal number all smaller cardinal numbers are *good* too).



$2$-colorings of triangles, resulting in $(2\ 2)$-colorings of all tetrahedra
When $|X|=7$, no such coloring exists. 


In this case there are 35 triangles and 35 tetrahedra. If $A$ is a tetrahedron, let $(\alpha\_A,\beta\_A)$ be its type. Each triangle in $X$ is a subset of exactly $4$ tetrahedra, so if 
$$\sum\_{T\subseteq X, |T|=3}c(T)=n$$
then
$$\sum\_{A\subseteq X, |A|=4} \beta\_A = 4n$$
If each $A$ were of type $(2,2)$, this would imply $4n=70$.


