= Solution
Define $D_{\mathcal G}(A)=\{x:A-x\in\mathcal G\}$. The proposed <addition of filters on the natural numbers> is
$$
\mathcal F+\mathcal G=\{A:D_{\mathcal G}(A)\in\mathcal F\}.
$$
Because addition of positive integers stays in $\mathbb N$, $D_{\mathcal G}(\mathbb N)=\mathbb N$ and $D_{\mathcal G}(\varnothing)=\varnothing$. Thus the sum contains the whole set and excludes the empty set. If $A\subseteq B$, upward closure of $\mathcal G$ gives $D_{\mathcal G}(A)\subseteq D_{\mathcal G}(B)$, so upward closure of $\mathcal F$ gives upward closure of the sum.
Finally, for every $x$,
$$
(A\cap B)-x=(A-x)\cap(B-x).
$$
The conjunction property for $\mathcal G$ proved in (i) consequently gives
$$
D_{\mathcal G}(A\cap B)=D_{\mathcal G}(A)\cap D_{\mathcal G}(B).
$$
Finite-intersection closure of $\mathcal F$ proves the same closure for its sum. All proper-filter axioms hold, so \b[(iv) is always true]. No ultrafilter assumption is needed here.
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