Baltic Way Shortlist, 2023

Since $F_n$ is monotonically increasing, it is clear that
$$0 < N < 2^{2022}F_{2023}.$$
Moreover, since $F_{n+1} \le 2F_n$, it is clear that $F_{2023} < 2^{2023}$ and hence $N$ has at most 4045 binary digits. It will thus suffice to prove that the last 2023 binary digits of $N$ are 1's, in other words, that $2^{2023} \mid N + 1$.
We have

Adding this three equations together, we get on the left hand side
$$4N - 2N - N = N$$
and on the right hand side
$$\begin{align*} &= 2^{2024} F_{2023} - 2^{2023} F_{2022} - 2^{2023} F_{2023} + \dots \ & \quad + \sum_{i=2}^{2022} (-2)^i (F_{i-1} + F_i - F_{i+1}) - 2F_1 - 2(-F_2) - F_1 \ &= 2^{2024} F_{2023} - 2^{2023} (F_{2022} + F_{2023}) + 2(F_2 - F_1) - F_1 \ &= 2^{2024} F_{2023} - 2^{2023} F_{2024} - F_1. \end{align*}$$
Hence
$$N + 1 = 2^{2024} F_{2023} - 2^{2023} F_{2024}$$
is indeed a multiple of $2^{2023}$ as desired.

For $n \in \mathbb{Z}_{\ge 2}$, let $N_n := \sum_{i=0}^{n} (-2)^i F_{i+1}$. As in the first proof, we want to show $2^{2023} \mid N_{2022} + 1$, or more generally $2^{n+1} \mid N_n + 1$. To do so, we prove the explicit representation of the sum as $N_n + 1 = (-1)^n \cdot 2^{n+1} \cdot F_{n-1}$.
For $n=2$, we have $N_2 + 1 = 2^2 F_3 - 2F_2 + F_1 + 1 = 8 = (-1)^2 \cdot 2^3 \cdot F_1$. The induction steps follows via
$$\begin{align*} N_{n+1} + 1 &= N_n + 1 + (-2)^{n+1} F_{n+2} \ &= -(-2)^{n+1} \cdot F_{n-1} + (-2)^{n+1} \cdot (F_{n+1} + F_n) \ &= (-2)^{n+1} \cdot (-F_{n-1} + F_n + F_{n-1} + F_n) = -(-2)^{n+2} \cdot F_n. \end{align*}$$
Thus, $2^{n+1} \mid N_n + 1$ and, in particular, $2^{2023} \mid N_{2022}$, which proves the claim. The desired results follows from there as in the first proof.