Example 1 Use the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once each to form two numbers, such that one number is twice the other. Find all solutions that satisfy the condition.
Show a four-digit number. Thus, the original number $M=2N$, or

as an odd number, there must be $D=5$, which then leads to $H=$ being 7. But $C=$, so $H=2, 2 \times H$ should not have a carry,
this contradiction indicates $C \neq 7$. (3) If $D=7$, knowing $H \neq 8$, we have $H=\frac{7-1}{2}=3$.
as an odd number, there must be $B=5$. Assuming there is a carry, $C$ is even, and $E$ is by $\overline{A B}=15=2 \times F+1$, we know $F=\frac{15-1}{2}=$

In $M$, 2,3 are in $N$. Conclusion 6 gives all 12 solutions.
$6729 \times 2=13458$ (solution)
$6792 \times 2=13584$ (solution), $6927 \times 2=13854$ (solution),
$\qquad$
$\qquad$ $7692 \times 2=15384$ (solution)
$7962 \times 2=15924$ (discard)
$9267 \times 2=18534$ (solution)
$9627 \times 2=19254$ (discard) $9672 \times 2=19344$ (discard)
$9726 \times 2=19452$ (discard) $9762 \times 2=19524$ (discard)
Take 7,
When $N$ contains $9,7,2,3$, there are 12 cases
$7923 \times 2=15846$ (solution),
$7392 \times 2=14784$ (discard)
$7239 \times 2=14478$ (discard) $7293 \times 2=14586$ (solution)
$9723 \times 2=19446$ (discard)
$9327 \times 2=18654$ (solution)
$9372 \times 2=18744$ (discard), $9237 \times 2=18474$ (discard),
$9273 \times 2=18546$ (solution)
$\qquad$ $6927 \times 2=13854, \quad 7269 \times 2=14538$, $7692 \times 2=15384, \quad 9267 \times 2=18534$,
$\begin{array}{ll}7923 \times 2=15846, & 7932 \times 2=15864, \\ 7329 \times 2=14658, & 7293 \times 2=14586,\end{array}$
$9327 \times 2=18654$, similarly solvable