Minimal cardinality of non-bipartite sub-family of $[\omega]^\omega$ Let $[\omega]^\omega$ the collection of infinite subsets of $\omega$. We say that $E\subseteq [\omega]^\omega$ is bipartite if there is $d\subseteq \omega$ such that for all $e\in E$ the intersections $e\cap d$ and $e\cap (\omega\setminus d)$ are both non-empty. If $E\subseteq[\omega]^\omega$ is countable, then $(\omega, E)$ is bipartite. (There is an elegant argument somewhere on MathOverflow for this by user @bof; I will add it as a comment once I find it.)
Now we can define the non-bipartiteness number by ${\frak nb} := \min\{|E|: E\subseteq [\omega]^\omega \text{ and } E \text{ is not bipartite}\}.$
Is it consistent that ${\frak nb}<{\frak c} = 2^{\aleph_0}$? Is it even consistent that ${\frak nb < a}$ where ${\frak a}$ is the minimum cardinality of a maximum almost disjoint family in $\omega$? (Only one question needs to be answered for acceptance.)
 A: The cardinal $\mathfrak{nb}$ is equal to the reaping number $\mathfrak{r}$.
An unsplit family is a collection $\mathcal R$ of infinite subsets of $\omega$ such that there is no set $D \subseteq \omega$ with the property that $E \cap D$ and $E \setminus D$ are infinite for every $E \in \mathcal R$. The reaping number $\mathfrak{r}$ is defined to be the minimum cardinality of an unsplit family.
Note that a bipartite family (from the question) is defined in almost exactly the same way: just replace "are infinite" with "are nonempty" in the definition of an unsplit family.
The point is that $\mathcal E$ is a bipartite family then it is also an unsplit family, and if $\mathcal R$ is an unsplit family then
$$\mathcal E = \{ X \subseteq \omega :\, X \Delta E \text{ is finite for some }E \in \mathcal R\}$$
is a bipartite family. Because neither kind of family can be finite, this implies that the minimum cardinality of a bipartite family, $\mathfrak{nb}$, is equal to the minimum cardinality of an unsplit family, $\mathfrak{r}$.
