1. Find the sum of all roots of the equation:

$$
\begin{gathered}
\sqrt{2 x^{2}-2024 x+1023131}+\sqrt{3 x^{2}-2025 x+1023132}+\sqrt{4 x^{2}-2026 x+1023133}= \\
=\sqrt{x^{2}-x+1}+\sqrt{2 x^{2}-2 x+2}+\sqrt{3 x^{2}-3 x+3}
\end{gathered}
$$

(L. S. Korechkova)
Solution. Note that the radicands in the left part are obtained from the corresponding radicands in the right part by adding $x^{2}-2023 x+1023130=$ $(x-1010)(x-1013)$. Since all radicands are positive (it is sufficient to check for $x^{2}-x+1$ and for $\left.2 x^{2}-2024 x+1023131=2(x-506)^{2}+511059\right)$, the left part is less than the right part when $1010<x<1013$ and greater when $x \notin[1010,1013]$. Equality is achieved only at $x=1010$ and $x=1013$, so the answer is -2023.

Criteria. If the answer is found but it is not proven that there are no other roots of the equation, no more than 2 points.