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Let $m$ be a given natural number and consider the additive group of rational numbers $\mathbb{Q}$. We are looking for a subset $A\subseteq \mathbb{Q}$ such that $A-A=\mathbb{Q}\setminus \{\pm1,\pm 2,\cdots,\pm m\}$.

For $m=1$, one can see the answer in Some co-finite subsets of rational numbers.

A candidate solution is of the form $A= \left\{ m+r_m+\sum_{k=m}^n r_k : n \ge m-1\right\} \cup B$, where $\Bbb{Q} \cap (m, +\infty) = \{ r_k \}_{k \ge m}$ and $B$ is an appropriate subset of $\Bbb{Q} \cap (-m,m)$ (but we are not sure about existence of such $B$).

Note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$.

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