I am trying to think to a possible answer to this question using generating functions built for a linear system of equations in binary variables (i.e. equal to $0$ or $1$) in the way described in this answer.
For the above cited question, let $x_1$ be the number of elements belonging to $S_1$ only, $x_{12}$ the number of elements belonging to $S_1$ and $S_2$ (and not to $S_3$), $x_{13}$ the number of elements belonging to $S_1$ and $S_3$ (and not to $S_2$), $x_{123}$ the number of elements belonging to $S_1$, $S_2$ and $S_3$ and so on. The condition that the three sets contain exactly $2k$ elements can be expressed through the following system:
$$\begin{cases} x_{1}+x_{12}+x_{13}+x_{123}=2k \\ x_{2}+x_{12}+x_{23}+x_{123}=2k \\ x_{3}+x_{13}+x_{23}+x_{123}=2k \\ \end{cases} $$
All the above variables are in $[0, 2k]$. We can express each of them as a sum of binary values ($0$ or $1$): $x_{1}=\sum_{i=1}^{2k}x_{i,1}$, $x_{2}=\sum_{i=1}^{2k}x_{i,2}$, $x_{3}=\sum_{i=1}^{2k}x_{i,3}$, $x_{12}=\sum_{i=1}^{2k}x_{i,12}$, etc.
The above system can be rewritten as:
$$\begin{cases} \sum_{i=1}^{2k}(x_{i,1}+x_{i,12}+x_{i,13}+x_{i,123})=2k \\ \sum_{i=1}^{2k}(x_{i,2}+x_{i,12}+x_{i,23}+x_{i,123})=2k \\ \sum_{i=1}^{2k}(x_{i,3}+x_{i,13}+x_{i,23}+x_{i,123})=2k \\ \end{cases} $$
Using the method described in the above cited answer, we can compute the number of solutions of this system as the coefficient of $z_1^{2k}z_2^{2k}z_3^{2k}$ in:
$$[(1+z_1)(1+z_2)(1+z_3)(1+z_1z_2)(1+z_1z_3)(1+z_2z_3)(1+z_1z_2z_3)]^{2k}$$
For example, for $k=2$, WolframAlpha gives $163373$.
Now let's go back to the cited question: we want to color the elements in red and green such that in any set there are exactly $k$ red elements and $k$ green elements. Let $x_1 = r_1 + g_1$, where $r_1$ are the red elements belonging to $S_1$ only and $g_1$ are the green elements belonging to $S_1$ only, and similarly $x_{12} = r_{12}+g_{12}$, and so on. We can write the constraints with the following system in binary variables:
$$\begin{cases} \sum_{i=1}^{k}(r_{i,1}+r_{i,12}+r_{i,13}+r_{i,123})=k \\ \sum_{i=1}^{k}(r_{i,2}+r_{i,12}+r_{i,23}+r_{i,123})=k \\ \sum_{i=1}^{k}(r_{i,3}+r_{i,13}+r_{i,23}+r_{i,123})=k \\ \sum_{i=1}^{k}(g_{i,1}+g_{i,12}+g_{i,13}+g_{i,123})=k \\ \sum_{i=1}^{k}(g_{i,2}+g_{i,12}+g_{i,23}+g_{i,123})=k \\ \sum_{i=1}^{k}(g_{i,3}+g_{i,13}+g_{i,23}+g_{i,123})=k \\ \end{cases} $$
And since that system can be split into two independent systems for the red elements and green elements respectively, the number of solutions of it can be computed as the square of the coefficient of $z_1^kz_2^kz_3^k$ in:
$$[(1+z_1)(1+z_2)(1+z_3)(1+z_1z_2)(1+z_1z_3)(1+z_2z_3)(1+z_1z_2z_3)]^{k}$$
For the $k=2$ case considered above, WolframAlpha gives $125$, and then a total of $125^2=15625$ solutions.
Assuming WolframAlpha is right, we have $15625 \lt 163373$ and it seems that most of the sets $S_1$, $S_2$, $S_3$ with exactly $4$ elements cannot have their elements colored in red and green so that each set has exactly $2$ red elements and $2$ green elements.
But intuitively, making some attempts, it is not easy to find such an example, rather I didn't find any. Therefore I think the above reasoning is wrong.
Someone can help me find where I am wrong?
