Let's determine the smallest and the largest multiple of 693 in which all ten digits appear exactly once. (The digit 0 cannot be the leading digit.)
The prime factorization of $693=3^{2} \cdot 7 \cdot 11$ indicates that the numbers we are looking for must be divisible by $3^{2}=9$, 7, and 11, and if these conditions are met, the numbers will also be divisible by their product.

Divisibility by 9 is ensured by the sum of the digits, because the sum of the ten digits, each taken once, is $45=9 \cdot 5$.

A number is divisible by 11 if and only if the difference between the sum of the digits in the odd positions (counting from the end) and the sum of the digits in the even positions is divisible by 11. This is because the place value of the $k$-th digit is $10^{k-1}$, and for odd $k$, $10^{k-1}-1$ is divisible by $10^{2}-1=99$, which is $11 \cdot 9$ (also true for $k=1$), and for even $k$, $10^{k-1}+1$ is divisible by $10+1$. Based on this, for numbers with at most four digits, the statement can be derived as follows: a number written with digits $A, B, C, D$ can be expressed as:

$$
\begin{gathered}
A \cdot 10^{3}+B \cdot 10^{2}+C \cdot 10+D= \\
=A\left(11 M_{1}-1\right)+B\left(11 M_{2}+1\right)+C\left(11 M_{3}-1\right)+D=11 M_{4}+(D+B)-(C+A)
\end{gathered}
$$

and the last form confirms our statement (where $M_{i}$ are integers).

In our case, denoting the two 5-digit sums by $S_{1}$ and $S_{2}$, we have $S_{1}-S_{2}=11 N$. Here, $N$ is odd because $S_{1}+S_{2}=45$, which is odd, and the sum and difference of two integers have the same parity. Only $N= \pm 1$ is possible, so one sum is 28 and the other is 17, because if $N= \pm 3$, the smaller sum would be 6, which is not possible with the sum of 5 different digits.

Divisibility by 7 will be ensured by direct trial.

To find the smallest multiple of 693, it cannot start with the smallest 4-digit number 1023, because this gives $3+0$ to $S_{1}$ and $2+1$ to $S_{2}$, and the remaining part of the 28 sum cannot be reached with the largest three digits: $9+8+7=24$. From the fact that the difference is 1, it follows that by increasing the first four digits to 1024, the sum of the digits in the odd positions can be uniquely achieved as 28, and the remaining digits for $S_{2}$ can also be filled. Thus, among the numbers that can be written, the smallest multiple of $9 \cdot 11=99$ is 1024375869, with the digits 7, 8, 9 in the odd positions and 3, 5, 6 in the even positions in ascending order.

This number is not divisible by 7, with a remainder of 6, but by swapping 9 and 8, the smallest increase that preserves divisibility by 11 can achieve divisibility by 7, because the increase is $(9-8)(100-1)=99=7 \cdot 14+1$. The smallest multiple is:

$$
1024375968=693 \cdot 1478176
$$

To find the largest multiple, we start with 9876. This gives $8+6=14$ to $S_{1}$, and the smaller sum of 17 can be achieved: $8+6+2+1+0=17$, and thus the largest multiple of $9 \cdot 11$ that can be written is 9876524130.

The remainder of this number when divided by 7 is 4. Swapping 1 and 0 would decrease the number by 99, changing the remainder to 3, which is not acceptable. Swapping the digits in the tens and hundreds places, 4 and 3, decreases the number by $(4-3)(1000-10)=990$, which is $7 \cdot 141+3$. Thus, performing both swaps, the remainder becomes 0, and the largest multiple is:

$$
9876523041=693 \cdot 14251837 .
$$

Csetényi Artúr (Kiskunhalas, Szűts J. Ált. Isk., 8. o. t. ) Pröhle Tamás (Budapest, Fazekas M. Gyak. G., II. o. t. )