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Yes. Partition $\omega$ into two disjoint infinite subsets $T_1$ and $T_2$. Recursively construct a $3$-uniform linear hypergraph (Steiner triple system) $\mathcal S\subset\binom\omega3$ so that each element of $\binom\omega2$ is contained in a unique element of $\mathcal S$, and each element of $\mathcal S$ meets both $T_1$ and $T_2$. Namely, we enumerate the elements of $\binom\omega2$, and when we come to an element $\{x,y\}$ of $\binom\omega2$ which is not already covered, choose $i\in\{1,2\}$ so that $\{x,y\}\not\subset T_i$ and choose a number $z\in T_i\setminus\{x,y\}$ which is not in any element of $\binom\omega3$ which has already been put into $\mathcal S$, and add the triple $\{x,y,z\}$ to $\mathcal S$.

Yes. Partition $\omega$ into two disjoint infinite subsets $T_1$ and $T_2$. Recursively construct a $3$-uniform linear hypergraph (Steiner triple system) $\mathcal S\subseteq\binom\omega3$ so that each element of $\binom\omega2$ is contained in a unique element of $\mathcal S$, and each element of $\mathcal S$ meets both $T_1$ and $T_2$. Namely, we enumerate the elements of $\binom\omega2$, and when we come to an element $\{x,y\}$ of $\binom\omega2$ which is not already covered, choose $i\in\{1,2\}$ so that $\{x,y\}\not\subseteq T_i$ and choose a number $z\in T_i\setminus\{x,y\}$ which is not in any element of $\binom\omega3$ which has already been put into $\mathcal S$, and add the triple $\{x,y,z\}$ to $\mathcal S$.

Yes. Partition $\omega$ into two disjoint infinite subsets $T_1$ and $T_2$. Recursively construct a $3$-uniform linear hypergraph (Steiner triple system) $\mathcal S\subset\binom\omega3$ so that each element of $\binom\omega2$ is contained in a unique element of $\mathcal S$, and each element of $\mathcal S$ meets both $T_1$ and $T_2$. Namely, we enumerate the elements of $\binom\omega2$, and when we come to an element $\{x,y\}$ of $\binom\omega2$ which is not already covered, choose $i\in\{1,2\}$ so that $\{x,y\}\not\subset T_i$ and choose a number $z\in T_i\setminus\{x,y\}$ which is not in any element of $\binom\omega3$ which has already been put into $\mathcal S$, and add the triple $\{x,y,z\}$ to $\mathcal S$.

P.S. More directly, consider the $3$-uniform hypergraph whose vertices are the nonempty finite subsets of $\omega$, and whose edges are the triples $\{x,y,z\}$ consisting of three distinct nonempty sets $x,y,z$ with symmetric difference $x\triangle y\triangle z=\varnothing$.

Equivalently, consider the $3$ uniform hypergraph whose vertices are the positive integers, and whose edges are triples $\{x,y,z\}$ of positive integers which are losing nim positions for the first player.

Yes. Partition $\omega$ into two disjoint infinite subsets $T_1$ and $T_2$. Recursively construct a $3$-uniform linear hypergraph (Steiner triple system) $\mathcal S\subset\binom\omega3$ so that each element of $\binom\omega2$ is contained in a unique element of $\mathcal S$, and each element of $\mathcal S$ meets both $T_1$ and $T_2$. Namely, we enumerate the elements of $\binom\omega2$, and when we come to an element $\{x,y\}$ of $\binom\omega2$ which is not already covered, choose $i\in\{1,2\}$ so that $\{x,y\}\not\subset T_i$ and choose a number $z\in T_i\setminus\{x,y\}$ which is not in any element of $\binom\omega3$ which has already been put into $\mathcal S$, and add the triple $\{x,y,z\}$ to $\mathcal S$.

Yes. Partition $\omega$ into two disjoint infinite subsets $T_1$ and $T_2$. Recursively construct a $3$-uniform linear hypergraph (Steiner triple system) $\mathcal S\subset\binom\omega3$ so that each element of $\binom\omega2$ is contained in a unique element of $\mathcal S$, and each element of $\mathcal S$ meets both $T_1$ and $T_2$. Namely, we enumerate the elements of $\binom\omega2$, and when we come to an element $\{x,y\}$ of $\binom\omega2$ which is not already covered, choose $i\in\{1,2\}$ so that $\{x,y\}\not\subset T_i$ and choose a number $z\in T_i\setminus\{x,y\}$ which is not in any element of $\binom\omega3$ which has already been put into $\mathcal S$, and add the triple $\{x,y,z\}$ to $\mathcal S$.

Yes. Partition $\omega$ into two disjoint infinite subsets $T_1$ and $T_2$. Recursively construct a $3$-uniform linear hypergraph (Steiner triple system) $\mathcal S\subset\binom\omega3$ so that each element of $\binom\omega2$ is contained in a unique element of $\mathcal S$, and each element of $\mathcal S$ meets both $T_1$ and $T_2$. Namely, we enumerate the elements of $\binom\omega2$, and when we come to an element $\{x,y\}$ of $\binom\omega2$ which is not already covered, choose $i\in\{1,2\}$ so that $\{x,y\}\not\subset T_i$ and choose a number $z\in T_i\setminus\{x,y\}$ which is not in any element of $\binom\omega3$ which has already been put into $\mathcal S$, and add the triple $\{x,y,z\}$ to $\mathcal S$.

P.S. More directly, consider the $3$-uniform hypergraph whose vertices are the nonempty finite subsets of $\omega$, and whose edges are the triples $\{x,y,z\}$ consisting of three distinct nonempty sets $x,y,z$ with symmetric difference $x\triangle y\triangle z=\varnothing$.

Equivalently, consider the $3$ uniform hypergraph whose vertices are the positive integers, and whose edges are triples $\{x,y,z\}$ of positive integers which are losing nim positions for the first player.