9. For any two distinct real numbers $x, y$, define $D(x, y)$ as the unique integer $d$ satisfying
$$
2^{d} \leqslant|x-y|<2^{d+1}
$$

The "scale" of $x \in \mathscr{F}$ in $\mathscr{F}$ is defined as the value of $D(x, y)$, where $y \in \mathscr{F}, x \neq y$.

Let $k$ be a given positive integer, and assume that each $x$ in $\mathscr{F}$ has at most $k$ different scales in $\mathscr{F}$ (these scales may depend on $x$). Find the maximum number of elements in $\mathscr{F}$.
9. The maximum number of elements in $\mathscr{F}$ is $2^{k}$.

For convenience, the "scale" between two real numbers $x$ and $y$ is still denoted as $D(x, y)$.

First, construct a set $\mathscr{F}$ with $2^{k}$ elements such that each element in $\mathscr{F}$ has at most $k$ different scales in $\mathscr{F}$. Let $\mathscr{F}=\left\{0,1, \cdots, 2^{k}-1\right\}$. The scale between any two numbers in $\mathscr{F}$ is in the set $\{0,1, \cdots, k-1\}$.

Second, prove that $2^{k}$ is the upper bound of the number of elements in $\mathscr{F}$. For any finite set $S$ of real numbers and any real number $x$, let $r_{S}(x)$ denote the number of different scales of $x$ in $S$, then
$r_{S}(x)=|\{D(x, y) \mid x \neq y, y \in S\}|$.
Thus, for each element $x$ in the set $\mathscr{F}$ that satisfies the condition, we have $r_{\mathscr{F}}(x) \leqslant k$.

Lemma: Let $S$ be a non-empty finite set of real numbers, and $w(S)=\sum_{x \in S} 2^{-r_{s}(x)}$, then $w(S) \leqslant 1$.

Proof: For $|S|=n \in \mathbf{Z}_{+}$, prove by mathematical induction.
If $n=1$, let $S=\{x\}$, then $r_{s}(x)=0$. Thus,
$w(S)=1$;
If $n=2$, let $S=\left\{x_{1}, x_{2}\right\}, x_{1}1$, for all $i, j$ with $1 \leqslant i < j \leqslant n$, we have $D\left(x_{i}, x_{j}\right)>d$, because
$$
\begin{array}{l}
\left|x_{i}-x_{j}\right|=\left|x_{i+1}-x_{i}\right|+\left|x_{j}-x_{i+1}\right| \\
\geqslant 2^{d}+2^{d}=2^{d+1} .
\end{array}
$$

Select the smallest positive integer $i \leqslant t$ and the largest positive integer $j \geqslant t+1$, such that
$$
\begin{array}{l}
D\left(x_{i}, x_{i+1}\right)=D\left(x_{i+1}, x_{i+2}\right)=\cdots \\
=D\left(x_{j-1}, x_{j}\right)=d .
\end{array}
$$

Let $E$ be the set of all $x_{l}$ where $l$ is even and $i \leqslant l \leqslant j$; $O$ be the set of all $x_{l}$ where $l$ is odd and $i \leqslant l \leqslant j$; $R$ be the set of the remaining elements in $S$.
Thus, $S$ is the union of $E$, $O$, and $R$.
Let $S_{O}=R \cup O, S_{E}=R \cup E$. Then
$\left|S_{O}\right|2^{k}$, then
$$
w(\mathscr{F})=\sum_{x \in \mathscr{F}} 2^{-r_{\mathscr{F}}(x)} \geqslant \sum_{x \in \mathscr{F}} 2^{-k}>1,
$$

which is a contradiction.
Therefore, $|\mathscr{F}| \leqslant 2^{k}$.
In summary, $|\mathscr{F}|_{\max }=2^{k}$.
(Provided by Xiong Bin, translated by Li Jianquan)