= Solution
Use $\mathbb N=\{1,2,\ldots\}$. If a <bijection> $f:\mathbb N\to\mathbb N$ maps $S$ onto the even <natural numbers>, its restriction is a <bijection> from $S$ onto an infinite <set>. Moreover, its restriction to the <complement> of $S$ maps onto the odd <natural numbers>. Both <sets> are therefore infinite.
Conversely, suppose that $S$ and its <complement> are infinite. Applying the <increasing enumeration of an infinite subset of natural numbers> to each gives
$$
S=\{s_1<s_2<\cdots\},\qquad\mathbb N\setminus S=\{t_1<t_2<\cdots\}.
$$
For completeness, successively choose the least unused member. The process never stops because the <set> is infinite, and it lists every member: an omitted member would have infinitely many selected <natural numbers> below it, which is impossible. Define
$$
\boxed{f(s_j)=2j,\qquad f(t_j)=2j-1.}
$$
This is <injective> on each list, their images are disjoint, and together their images cover $\mathbb N$. Thus it is a <bijection> and maps $S$ onto $2\mathbb N$. \b[Both infinitude conditions are necessary and sufficient.] If a convention includes zero in $\mathbb N$, index each list from zero and use $2j$, $2j+1$ instead.
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