Fix a positive integer $m$ and put $$A_m:= \{ \sum_{k=m+1}^n k: n \ge m\}=\{\frac{1}{2}n^2+\frac{1}{2}n-\frac{1}{2}m^2-\frac{1}{2}m: n \ge m\}= \{\frac{1}{2}(n-m)(n-m+1): n \ge m\} $$ It is not difficult to show that $(A_m-A_m)+\{0,1,\cdots,m \} = \Bbb{Z}$ and $(A_m-A_m)\cap (\{0,1,\cdots,m \}-\{0,1,\cdots,m \})=\{0\}$ (because $A_m-A_m = \Bbb{Z} \setminus \{ \pm 1,\cdots,\pm m \}$).
Now, is it true that if $B\subseteq \Bbb{Z}$ has the properties $(A_m-A_m)+B= \Bbb{Z}$ and $(A_m-A_m)\cap (B-B)=\{0\}$, then $|B|=m+1$ (i.e., every subset $B$ satisfying the conditions is as size as $\{0,1,\cdots,m \}$).
Note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$.
This is true: writing for brevity $D=A_m-A_m=\mathbb Z\setminus\{\pm1,\dotsc,\pm m\}$, if $D+B=\mathbb Z$ and $D\cap(B-B)=\{0\}$, then $|B|=m+1$; moreover, in this case $B$ is a block of consecutive integers of length $m$.
To see this, notice first that the condition $D\cap(B-B)=\{0\}$ implies $\max B-\min B\le m$; that is, $B$ is contained in an interval of length $m$. Without loss of generality, we can assume that $\min B=0$, and then $0\in B\subseteq[0,m]$. Indeed, for each $k\in[1,m]$ we must have $k\in B$, since $k\in[0,m]\setminus B$ is easily seen to imply $k\notin D+B$ (the sets $-D$ and $B-k$ are disjoint showing that $0\notin D+B-k$, whence $k\notin D+B$).