Bulgarian National Olympiad - Final Round

Suppose that such $a_i$ exist. The condition rewrites as ${\sum }_{i=1}^{2024}{\left(a_i+\frac{1}{a_{i+1}}-1\right)}^2=0$, which means that $a_i+\frac{1}{a_{i+1}}=1$ for all $i=1,2,\dots ,2024$. It's easy to obtain consecutively that for any $i=1,2,\dots ,2024$, $a_{i+1}=\frac{1}{1-a_i}$, $a_{i+2}=1-\frac{1}{a_i}$, so $a_{i+3}=a_i$. Since $3\nmid 2024$, all $a_i$ are equal to some real $k$. However, $k+\frac{1}{k}=1⟺k^2-k+1=0$ doesn't have a real solution, contradiction. $\square$