= 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 \b[no such <ultrafilter> exists].
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