74th Romanian Mathematical Olympiad

If $n\in \mathbb{N}$, then $a_1{n}^{1012}+a_2{n}^{1011}+\cdots +a_{1012}n+a_{1013}>0$, hence $n\in \mathbb{N}$ is not suitable.

If $n=-k$, with $k\ge 2$, then $a_1{n}^{1012}+a_2{n}^{1011}+\cdots +a_{1012}n+a_{1013}=k^{1011}(a_{1}k-a_2)+k^{1009}(a_{3}k-a_4)+\cdots +k(a_{1011}k-a_{1012})+a_{1013}>0$, because $a_{1013}>0$ and $a_{i}k-a_{i+1}\ge 2a_i-a_{i+1}\ge 21012-2024=0$, for every $i\in \{1,2,\dots ,1011\}$. Hence, there are no solutions of this type.

We show that $n=-1$ is a solution. For each $a\in \mathbb{Z}$, $(a-1)-a-(a+1)+(a+2)=0$. It follows that $(1012-2024+1013-1015+1014)+(1016-1017-1018+1019)+\cdots +(2020-2021-2022+2023)=0$, hence $n=-1$ is the only solution.