Problem 1700

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 $A_iA_{i+1}$ 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 $A_iA_{i+1}$ 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:
$$(1, 1, -2), (1, 1, 2), (-1, -1, 2), (-1, -1, -2), \\ (2, 2, -1), (2, 2, 1), (-2, -2, 1), (-2, -2, -1).$$
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. $\square$