= 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. Consequently
$$
\{\mathbb N\setminus B:B\text{ is bad}\}
$$
has 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 $n$ by
$$
v_2(n)+v_3(n)\pmod2.
$$
Multiplication 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.
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