 Solution

Every ultrafilter contains exactly one cell of any finite set partition. Indeed, two disjoint cells cannot both belong to a proper filter, while if none belonged to the ultrafilter, all their complements would belong to it and their finite intersection would be empty.

Apply this property to the two colour classes in (iii). One belongs to any proposed ultrafilter, but neither contains a symmetric weighted-pair configuration from an increasing infinite sequence. This contradicts the proposed property of every member. Therefore no such ultrafilter exists.