For 10 positive integers $x$, let $m=\left[\frac{x}{2}\right]+\left[\frac{x}{2^{2}}\right]+\left[\frac{x}{2^{3}}\right]+\cdots+\left[\frac{x}{2^{k}}\right]$, where $k$ is the smallest integer satisfying $2^{k} \geqslant x$, and the symbol $[x]$ represents the greatest integer not exceeding $x$. The difference between $x$ and $m$, i.e., $x-m$, is called the "deficiency number" of the positive integer $x$. (For example, when $x=100$,
\[
\begin{aligned}
m & =\left[\frac{100}{2}\right]+\left[\frac{100}{2^{2}}\right]+\left[\frac{100}{2^{3}}\right]+\left[\frac{100}{2^{4}}\right]+\left[\frac{100}{2^{5}}\right]+\left[\frac{100}{2^{6}}\right]+\left[\frac{100}{2^{7}}\right] \\
& =50+25+12+6+3+1+0=97,
\end{aligned}
\]
$x-m=100-97=3$, so the "deficiency number" of the number 100 is 3. ) Then the smallest positive integer $x$ whose "deficiency number" is 9 is $\qquad$
(10) 511 Prompt: Let the binary representation of a positive integer $x$ be
$$
x=a_{n} \times 2^{n}+a_{n-1} \times 2^{n-1}+\cdots+a_{2} \times 2^{2}+a_{1} \times 2^{1}+a_{0} \times 2^{0},
$$

where $a_{i}=0$ or $1(i=0,1,2, \cdots, n)$, and $a_{n} \neq 0$.
It is easy to see that,
$$
\begin{array}{c}
x-2 \times\left[\frac{x}{2}\right]=a_{0}, \\
{\left[\frac{x}{2}\right]-2 \times\left[\frac{x}{2^{2}}\right]=a_{1},} \\
{\left[\frac{x}{2^{2}}\right]-2 \times\left[\frac{x}{2^{3}}\right]=a_{2},} \\
\cdots \\
{\left[\frac{x}{2^{n}}\right]-2 \times\left[\frac{x}{2^{n+1}}\right]=a_{n} .}
\end{array}
$$

Therefore,
$$
\begin{aligned}
& a_{0}+a_{1}+a_{2}+\cdots+a_{n} \\
= & x-2 \times\left[\frac{x}{2}\right]+\left[\frac{x}{2}\right]-2 \times\left[\frac{x}{2^{2}}\right]+\left[\frac{x}{2^{2}}\right] \\
\cdot & -2 \times\left[\frac{x}{2^{3}}\right]+\cdots+\left[\frac{x}{2^{n}}\right]-2 \times\left[\frac{x}{2^{n+1}}\right] \\
= & x-m .
\end{aligned}
$$

Thus, the "deficiency number" of the number $x$ is $a_{0}+a_{1}+a_{2}+\cdots+a_{n}$, which is the number of non-zero digits in the binary representation of $x$.

Therefore, the smallest positive integer with a deficiency number of 9 is $1+2^{1}+2^{2}+\cdots+2^{8}=2^{9}-$ $1=511$