For the digits of the decimal number $\overline{a b c d}$, it holds that $a>b>c>d$. These same digits, in some order, are also the digits of the difference $\overline{a b c d}-\overline{d c b a}$. Which is this four-digit number?
When the subtraction is written in the usual scheme, in the last two columns (1) the lower digit is larger than the upper one; here, therefore, we exchange one ten of the minuend for units, or one hundred for tens, so the difference in the units and tens place value is:

$$
\begin{array}{ll}
\text { units: } & e=10+d-a=10-(a-d) \\
\text { tens: } & t=10+(c-1)-b=9-(b-c)
\end{array}
$$

However, in the first two columns, the upper digit is larger, so the other digits of the difference are:

$$
\begin{aligned}
& \text { hundreds: } \quad s=(b-1)-c=(b-c)-1, \\
& \text { thousands: } \quad k=a-d
\end{aligned}
$$

( $k$ as the first letter of "kilometer"). We see that

$$
k+e=10 \text { and }
$$

$$
\text { (7) } \quad s+t=8
$$

We can also determine the magnitude relationships for the last two and the first two digits of the difference based on (1):

$$
t=9-b+c \geq 9-(a-1)+(d+1)=e+1>e, \quad k=a-d>a-c>b-c>s
$$

Furthermore, $k, s, t, e$ represent the original digits in some order, so from the above, we conclude that the smallest digit, $d$, is either $e$ or $s$ in the difference, and the digit $a$ is either $k$ or $t$.

However, the assumption $d=e$ leads to a contradiction: from (2) it follows that $a=10$; therefore, the smallest digit can only be $d=s$.

Knowing this, the assumption $a=k$, i.e., $a>t$, also leads to a contradiction: from (5) it follows that $d=s=0$, from (7) $t=8$, so clearly $a=k=9$ and from (6) $e=1$ follows; the only possible values for the remaining digits are 8 for $b$ and 1 for $c$, but then the condition $b-c=1$ from (4) is not satisfied. Therefore, the largest digit can only be $a=t$, and from (7) $a \leq 8$, and if we do not allow $d=0$ as the leading digit, then $a \leq 7$.

Then $b$ and $c$ are one $k$ and one $e$, and from (6) $b+c=10$, so $b$ and $c$ have the same parity; therefore, their difference is also even: $b-c \geq 2$, so $c \leq 4$ and $b \geq 6$. From this, $a \geq 7$; and if we accept the restriction $d \geq 1$, then only

$$
a=7, \quad d=1, \quad b=6, \quad c=4
$$

is possible. This is also a solution to the problem, and the sought number is 7641.

Starting from $d=0$ would yield $a=8, e=2=c$ and $b=8=a$, so there is no other solution.