Note that for $n=2$, since no two consecutive integers are equal, no such sets exist. So $n \geq 3$. Also, if $\{a+1, a+2, \ldots, a+n\}$ works, then $\{-a-1, -a-2, \ldots, -a-n\}$ works as well. WLOG assume that the set $\{a+1, a+2, \ldots, a+n\}$ consists of only nonnegative integers. The greatest element must be greater than or equal to all other elements, and the set contains at least $3$ elements, so $a+n \geq a+(n-1)+a+(n-2) \implies a+n \leq 3$. Since $a \geq -1$ and $n \geq 3$, $(a, n) = (-1, 3), (-1, 4), (0, 3)$. $(a, n) = (-1, 3)$ gives us $\{0, 1, 2\}$, which doesn't work, but $(a, n) = (-1, 4), (0, 3)$ give us $\boxed{\{0, 1, 2, 3\}}$ and $\boxed{\{1, 2, 3\}}$ respectively, which both work. Similarly, we see that $\boxed{\{-3, -2, -1, 0\}}$ and $\boxed{\{-3, -2, -1\}}$ work as well. Now assume that the set contains both positive and negative integers. Note that if the set has the same number of positive and negative integers, then $0$ is equal to the sum of all other elements. Thus, $\boxed{\{-k, -k+1, \ldots, k-1, k\}}$ works for all positive integers $k$. WLOG assume that there are more positive integers than negative integers in the set. It becomes $\{-n, -n+1, \ldots, n+m-1, n+m\}$ where $n, m \geq 1$. If $m=1$, then the sum of all the elements is $n+1$ which must be even. Thus $n$ must be odd. This gives us $\boxed{\{-2k+1, -2k+2, \ldots, 2k\}}$ for all positive integers $k$. If $m=2$, then the sum of all the elements is $2n+3$, which is odd, a contradiction. If $m \geq 3$, since the greatest element must be greater than or equal to all other elements, $m+n \geq m+n-2+m+n-1 \implies m+n \leq 3$. Since $n \geq 1$, we have a contradiction. Similarly, we can see that $\boxed{\{-2k, -2k+1, \ldots, 2k+1\}}$ works for all positive integers $k$. Those are all the sets that satisfy this property.