No.

Note that $\left(a_i + \frac{1}{a_{i+1}}\right)^2 = a_i^2 + 2\frac{a_i}{a_{i+1}} + \frac{1}{a_{i+1}^2}$, summing over all indices modulo $2024$ gives \[\sum \left(a_i + \frac{1}{a_{i+1}}\right)^2 - 2 \sum \left(a_i + \frac{1}{a_{i+1}}\right) + \sum 1 = 0\]Hence $a_i + \frac{1}{a_{i+1}} = 1$ for all $i$. Let $a_1=a$, $a_2 = \frac{1}{1-a}$, $a_3 = \frac{a-1}{a}, a_4 = a$ so the sequence is periodic modulo $3$ and the entire sequence can be characterized this way. In the end, $a_{2024} = a_2 = \frac{1}{1-a}$ and $a_{2024} + \frac{1}{a_1} =1$ which can be verified to have no real solutions.

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