Bulgarian Spring Tournament

Clearly, if there are two adjacent points $A_i$ and $A_{i+1}$ with opposite numbers, the problem is solved. Without restriction, let $A_1=1$ and consider all segments $A_i{A}_{i+1}$ for $i=1,2,\dots ,1012$. Since $A_{1013}=-1$, among the considered segments there is an odd number whose ends are one positive and one negative number. These two numbers can be $\{-1,2\}$ or $\{1,-2\}$ and let the number of segments with ends of the first kind be $p$ and the number of segments with ends of the second species to be $q$. Due to symmetry, the number of segments $A_i{A}_{i+1}$ for $i=1013,\dots ,2024$ ($A_{2025}\equiv A_1$) with ends $\{1,-2\}$ is equal to $p$. Therefore, all segments with endpoints $\{1,-2\}$ are $p+q$, which is an odd number. Now consider any triangle that does not have two vertices with opposite numbers. We have the following possibilities for the three numbers: $$\begin{gathered} (1,1,-2),(1,1,2),(-1,-1,2),(-1,-1,-2), \\ (2,2,-1),(2,2,1),(-2,-2,1),(-2,-2,-1). \end{gathered}$$ Each of these triangles has an even number (2 or 0) of sides with ends 1 and $-2$. Therefore, the total number of segments with ends 1 and $-2$ (counted in multiples) is an even number. But every line segment inside the 2024-gon is counted twice (once from the two triangles in which it participates), and every line segment that is a side of the 2024-gon is counted once. The resulting contradiction shows that a triangle with the requested property exists. $\square$