 Solution

Assume condition i and call a set $B\subseteq \mathbb{N}$ bad when it contains no member of $\mathcal{S}$. No finite collection of bad sets covers $\mathbb{N}$: if it did, assigning each integer to the first bad set containing it would give a finite coloring whose color classes are bad, contrary to condition i. Consequentlyhas the finite-intersection property. It generates a proper filter on a set, which the ultrafilter lemma extends to an ultrafilter $\mathcal{U}$. If some $A\in \mathcal{U}$ were bad, then $\mathbb{N}\setminus A$ would also belong to $\mathcal{U}$ by construction, contradicting propriety. Hence every $A\in \mathcal{U}$ contains a member of $\mathcal{S}$, proving condition ii.

For the final question, let $\mathcal{S}$ consist of the pairs $\{x,2x\}$ and $\{x,3x\}$. Color $n$ byMultiplication by either two or three reverses this parity, so this two-coloring has no monochromatic member of $\mathcal{S}$. Condition i fails, and the equivalence just proved shows that no ultrafilter with the stated property exists.