12. There is a four-digit number $A$, rearranging the digits of the four-digit number (none of which are 0) to form the largest possible number is 7668 greater than $A$, and the smallest possible number is 594 less than $A$, then $A=$ $\qquad$ .
【Analysis】First, we know the difference between $A$ and the maximum and minimum values, which means we know the difference between the maximum and minimum numbers. So, we need to find the relationship between the maximum and minimum values, $7668+594=8262$. Then, based on the positions of the maximum and minimum numbers being opposite, we set up a vertical arithmetic puzzle. Subtracting 7668 from the maximum number will give us the $A$ we are looking for.

【Solution】Solution: Let the largest four-digit number be $\overline{\mathrm{abcd}}$, and the smallest four-digit number be $\overline{\mathrm{dcba}}$.
The difference between the maximum and minimum is $7668+594=8262$
Expressed in a vertical format:
$$
\begin{array}{r}
a b c d \\
-\quad d c b a \\
\hline 8262
\end{array}
$$

First, we determine that $a=9, d=1$,
$b>c$, and at the tens place, $c-b$ is insufficient and requires borrowing. This gives us $b-c=3$. This condition is satisfied.
When $c=1$, $b=4$. This satisfies the vertical format. The maximum number is $9411-7668=1743$, which does not meet the condition.
When $c=2$, $b=5$. This satisfies the vertical format. The maximum number is $9521-7668=1853$, which does not meet the condition.
When $c=3$, $b=6$. This satisfies the vertical format. The maximum number is $9631-7668=1963$, which meets the condition.
Therefore, the answer is: 1963