Saudi Arabian Mathematical Competitions

The greatest element of set $A_{n}$ is
$$1+2+\ldots+n=\frac{n(n+1)}{2}$$
and the smallest element of $A_{n}$ is
$$-1-2-\ldots-n=-\frac{n(n+1)}{2} .$$
Also, the difference of any two elements of $A_{n}$ is even, hence all elements of $A_{n}$ are of the same parity.
Let us prove that all integers between $-\frac{n(n+1)}{2}$ and $\frac{n(n+1)}{2}$, and of the same parity with $\frac{n(n+1)}{2}$, belong to $A_{n}$, and these are all the elements of $A_{n}$. Indeed, let $x \in A_{n}$ be an element such that $-\frac{n(n+1)}{2} \leq x<\frac{n(n+1)}{2}$.
Case 1. If the writing of $x$ begins with -1 , then by changing -1 by +1 , we get $x+2 \in A_{n}$.
Case 2. If the writing of $x$ begins with +1 , then consider the first term with sign -. A such term exists, otherwise
$$x=1+2+\ldots+n=\frac{n(n+1)}{2} .$$
In this case we have
$$x=+1+2+\ldots+(k-1)-k \pm \ldots \pm n,$$
where $-k$ is the considered term. By changing the signs of terms $k-1$ and $k$, it follows $x+2 \in A_{n}$.
We obtain
$A_{n}=\left\{-\frac{n(n+1)}{2},-\frac{n(n+1)}{2}+2, \ldots, \frac{n(n+1)}{2}-2, \frac{n(n+1)}{2}\right\}$, that is the number of elements in $A_{n}$ is $\frac{n(n+1)}{2}+1$.