For every integer $k \geqslant 0$, we denote $F_{k}$ as the $k^{\text {th }}$ Fibonacci number, defined by $F_{0}=0$, $F_{1}=1$, and $F_{k}=F_{k-2}+F_{k-1}$ when $k \geqslant 2$. Let $n \geqslant 2$ be an integer, and let $S$ be a set of integers with the following property:

For every integer $k$ such that $2 \leqslant k \leqslant n$, the set $S$ contains two integers $x$ and $y$ such that $x-y=F_{k}$.
What is the smallest possible number of elements in such a set $S$?
First, let $m=\lceil n / 2\rceil$, and let $S$ be the set $\left\{F_{2 \ell}: 0 \leqslant \ell \leqslant m\right\}$. For any integer $k$ such that $2 \leqslant k \leqslant n$, we choose $x=F_{k}$ and $y=F_{0}=0$ if $k$ is even, or $x=F_{k+1}$ and $y=F_{k-1}$ if $k$ is odd. In both cases, $x$ and $y$ are two elements of $S$ such that $x-y=F_{k}$.
Conversely, let $S$ be a set as described in the statement. We will prove that $|S| \geqslant m+1$. To do this, we construct a weighted, undirected graph $G$ as follows. The vertices of our graph are the elements of the set $S$. Then, for every odd integer $k$ such that $1 \leqslant k \leqslant n$, and recalling that $F_{1}=F_{2}$, we choose two elements $x$ and $y$ of $S$ such that $x-y=F_{k}$. We then insert the edge $\{x, y\}$ into our graph $G$, and assign it the weight $F_{k}$.
The graph thus obtained has $m$ edges. Furthermore, suppose it contains a cycle $c=x_{0} x_{1} \ldots x_{\ell}$, with $x_{0}=x_{\ell}$, which we choose to have minimal length $\ell$. Without loss of generality, we assume that $x_{0}>x_{1}$, and that $\left\{x_{0}, x_{1}\right\}$ is the edge of $c$ with the maximum weight. We denote this weight by $F_{2 k+1}$. The cumulative weight of the other edges does not exceed $F_{1}+F_{3}+\ldots+F_{2 k-1}$ and, by construction, it is equal to

$$
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{\ell-1}-x_{\ell}\right| \geqslant\left|x_{\ell}-x_{1}\right|=F_{2 k+1}
$$

However, a straightforward induction on $k$ shows that $F_{1}+F_{3}+\ldots+F_{2 k-1}=F_{2 k}<F_{2 k+1}$. Consequently, our graph is acyclic, and since it has $m$ edges, it has at least $m+1$ vertices.
The minimal cardinality sought is therefore equal to $m+1$.