Suppose $A \subseteq \mathbb N^+$ is any arbitrary set. I want to determine under which conditions some finite set of translates $\{A + k_i : 1 \leq i \leq n\}$ of $A$ cover almost all of $\mathbb N^+$, i.e. when there is a finite set of natural numbers $\{k_i : 1 \leq i \leq n\}$ such that $$\mathbb N^+ \setminus \bigcup_{i=1}^n (A + k_i)$$ is finite.
It seems to me that a natural indicator of this possibility is the natural density of $A$ being positive; but perhaps it also suffices that the upper density is positive, or that the lower density is positive.
Question: Is it the case that a finite set of translates of $A$ covers almost all $\mathbb N^+$ if $A$ has positive upper density? If not, does positive lower density suffice?
Let $A$ be the set of numbers that are not of the from $m!+r$ with $0\le r<m$. Then $A$ has (lower/upper/normal) density $1$, but more importantly, it has arbitrarily large gaps. Hence $\bigcup (A+k_i)$ will not completely cover any gap of size $>\max\{k_1,\ldots, k_n\}$.
This example ledas us to
Theroem. $A$ has the "finite-translation-cover property" if and only if the gaps in $A$ are bounded.
Proof. If the gaps are not bounded, the the same argument as for the concrete example applies. On the other hand, if all gaps are $\le N$, then the translates by $1,2,\ldots,N$ cover almost all of $\Bbb N$. $\square$