We adopt the following convention: when writing a positive integer as a sum of Fibonacci numbers, we use the largest possible Fibonacci numbers. For example, although $18$ can be written as $8+5+5$, we write it as $13+5$.
Let $k$ be a positive integer and let $\mathcal{F}$ be the set of Fibonacci numbers. Define set addition $A+B := \{a+b: a \in A, b \in B\}$. Let $N(k)$ be the length of the longest arithmetic progression contained in the set $\underbrace{\mathcal{F} + \dots + \mathcal{F}}_{k}$.
We will prove by mathematical induction that $N(k)$ exists and is finite for every $k$; then taking $N = N(2023) + 1$ gives what we want.

Base case: when $k=1$, $N(1) \le 4$, because the longest arithmetic progression in $\mathcal{F}$ is $0,1,2,3$, of length $4$.

Next we prove two lemmas.

Lemma 1. If $a > F_n$, and $a$ can be written as a sum of $k$ terms $F_i$, each satisfying $F_i \le F_n$, then $a$ can also be written as a sum of $k$ terms $F_i$ satisfying $F_i \le F_n$, among which $F_n$ appears.

Proof of Lemma 1. Let $a = \sum_{j=1}^{k} F_{a(j)}$. Suppose $a$ cannot be written in such a form. Note that $2F_t = F_{t+1} + F_{t-2}$. If
- among the $F_{a(j)}$ there are two consecutive Fibonacci numbers $F_i + F_{i+1}$, we change them to $F_{i+2} + F_0$.
- among the $F_{a(j)}$ there are two repeated Fibonacci numbers $F_i + F_i$, and these two terms are not $0$, we change them to $F_{i+1} + F_{i-2}$.
We keep performing these two operations. If in this process $F_n$ never appears, then we can write $a$ as $\sum_{j=1}^{k} F_{b(j)}$, with $F_{b(j)} < F_n$, and, apart from $0$, there are no repeated or consecutive Fibonacci numbers. The maximum value of such a number is
$$F_{n-1} + F_{n-3} + F_{n-5} \le F_n < a,$$
a contradiction. Hence $a$ can be written as a sum of $k$ Fibonacci numbers each at most $F_n$, among which $F_n$ appears. □

In particular, if $F_n < a < F_{n+1}$, and $a$ can be written as a sum of $k$ Fibonacci numbers, then we may require that one of them be $F_n$.

Lemma 2. Given a positive integer $M$, then for any arithmetic progression of positive integers of length $4M$
$$A, A+d, \dots, A+(4M-1)d,$$
there exists an integer $\ell$ such that at least $M$ terms of this arithmetic progression lie in the interval $[F_\ell, F_{\ell+1}]$.

Proof of Lemma 2. The case $d=0$ needs no proof, so assume $d>0$. If $A+(3M)d$ through $A+(4M-1)d$ all lie in some interval $[F_\ell, F_{\ell+1}]$, the conclusion holds. Otherwise, there is some $F_\ell$ that cuts through this range, so we may write $A+md \le F_\ell < A+(m+1)d$, where $m \ge 3M$.
Note that $A+(3M)d \ge F_3 = 1$. And when $\ell \ge 3$, we have $1.5F_{\ell-1} = F_{\ell-1} + \frac{1}{2}(F_{\ell-2} + F_{\ell-3}) \le F_\ell$. Therefore from $1.5F_{\ell-1} \le F_\ell < A+(m+1)d$, we can obtain
$$F_{\ell-1} < \frac{A}{1.5} + \frac{m+1}{1.5}d \le A + (m+1-M)d,$$
so the $(m+1-M)$-th term through the $m$-th term all lie in the interval $[F_{\ell-1}, F_\ell]$. □

Remark: The estimate in this part is not the tightest possible; for instance, $1.5$ can be replaced by any $r$ between $1$ and $\frac{1+\sqrt{5}}{2}$, and $4$ can be replaced by any number $L$ satisfying $\frac{L-1}{r} > L-2$.

Finally we prove that $N(k) < 4N(k-1)+4$. By the induction hypothesis $N(k-1)$ is finite, so for any arithmetic progression of length $4N(k-1)+4$, there exists a sub-progression $a_1, a_2, \dots, a_{N(k-1)+1}$ of length $N(k-1)+1$ lying in some interval $[F_\ell, F_{\ell+1}]$. By Lemma 1 we know that if the $a_i$ can all be written as sums of $k$ Fibonacci terms, then we may require that $F_\ell$ appear among these $k$ terms. Thus $\langle a_i - F_\ell \rangle$ is an arithmetic progression of length $N(k-1)+1$, each of whose terms can be written as a sum of $k-1$ Fibonacci numbers, contradicting the definition of $N(k-1)$. Hence $N(k)$ is at most $4N(k-1)+3$. This completes the proof. □