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There are three sets, each of which contains some $2 k$ elements. The sets may overlap.

Is it possible to color the elements red and green, such that each set contains exactly $k$ red and $k$ green elements?

If $k$ is odd, then this may be impossible. For example, suppose the sets are: $$ S_1 = \{1,2,~~ 4,5,~~ 7,8,~~ \ldots, 3k-2, 3k-1 \} \\ S_2 = \{2,3,~~ 5,6,~~ 8,9,~~ \ldots, 3k-1, 3k \} \\ S_3 = \{3,1,~~ 6,4,~~ 9,7,~~ \ldots, 3k, 3k-2 \} $$ Each set contains exactly $2k$ elements. Each element appears in exactly two sets, so if there were exactly $k$ red elements in each set, the number of red elements overall should have been $3 k /2$ by double-counting. But this is impossible since $k$ is odd.

The remaining case is that $k$ is even. Is it always possible to split the sets in this case?

Erel Segal-Halevi
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    I would write a system like this: $$\begin{cases} x_{1}+x_{12}+x_{13}+x_{123}=2k \ x_{2}+x_{12}+x_{23}+x_{123}=2k \ x_{3}+x_{13}+x_{23}+x_{123}=2k \
    \end{cases} $$ where e.g. $x_{12}$ are the elements belonging to $S_1$ and $S_2$ only, and so on. Then divide each variable in equal parts for the red and green elements. For odd values, we are left with $1$ element unpaired. We rewrite the system $\pmod 2$, then we have only $5$ or $6$ isomorphic classes of solutions (surely less than $10$). We can then color the elements appropriately for each of these specific solutions.     – Fabius Wiesner Mar 28 '22 at 14:06
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    And probably the only case that we cannot color is the one that you found: $x_1=x_2=x_3=x_{123}=0, x_{12}=x_{13}=x_{23}=1$. – Fabius Wiesner Mar 28 '22 at 14:32

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