Is there a number in which every non-zero digit is used exactly once, and for which the number formed by the first $k$ digits is divisible by $k$ for every $k$ between 1 and 9?
The sought number is a nine-digit number. The condition for $k=1$ does not impose a restriction. The same is true for $k=9$, since the sum of the first nine positive integers is 45, and according to the divisibility rule for 9, any number that meets the conditions is divisible by 9. For brevity, the number formed by the first $k$ digits of the number will be referred to as the $k$-th slice of the number.

The fifth slice is divisible by 5 if it ends in 0 or 5. Since we cannot use 0, the fifth digit can only be 5.

The even slices, according to the condition, are even numbers, so in the sought number, the even positions are occupied by the four even digits. For $k=2$, this is already sufficient, but for $k=4,6,8$, additional conditions are also necessary.

It is known that a number is divisible by 4 if the last two digits, and by 8 if the last three digits form a number that is divisible by 4, respectively 8. In the latter case, however, an even number stands in the hundreds place - this is the sixth digit of the sought number - so the eighth slice will be divisible by 8 if and only if the two-digit number formed by the seventh and eighth digits is also divisible by 8.

In the sought number, odd numbers stand in the odd positions, so according to the above, a number divisible by 4 cannot stand in the fourth and eighth positions.

The third slice is divisible by 3 if and only if the sum of the first three digits, and since the sixth digit is even, the sixth slice is divisible by 6 if the sum of the first six digits is divisible by 3. The two conditions are simultaneously satisfied if and only if the sum of both the first and the second three digits is divisible by 3.

We have not yet used the condition for divisibility by 7. Based on the above, we distinguish two cases according to which of the two even numbers divisible by 4 stands in the sixth position.

**Case I.** The sixth digit is 8. Then the 4 is in the second position. The fifth digit is 5, so for the sum of the second three digits to be divisible by 3, we must choose the 2 from the 2 and 6 for the fourth position. Then the eighth digit is 6.

So in this case, we know the following about our number:

.4 .258 .6\).

Let's examine where the remaining four odd numbers can go! The 4 in the second position gives a remainder of 1 when divided by 3, and since none of the numbers 1, 3, 7, 9 give a remainder of 2 when divided by 3, an odd number divisible by 3 cannot stand in the first or third position if we want the sum of the first three digits to be divisible by 3.

The eighth digit is 6, so for the number formed by the seventh and eighth digits to be divisible by 8, the seventh position must be 9. Then the last digit is 3.

So we have two possibilities: 147258963 and 741258963. Now only the condition for $k=7$ remains, and it can be easily seen that it is not satisfied in either case.

**Case II.** The sixth digit is 4. Then the positions of the even numbers can be determined similarly to Case I, and we arrive at the following result:

$$
.8 .654 .2
$$

The divisibility by 3 now does not exclude any of the remaining odd numbers from the first or third position, we can only say that one of the two positions must be occupied by a number divisible by 3. The eighth digit is 2, so the eighth slice can be divisible by 8 in two ways, with 3 or 7 in the seventh position. In both cases, we have four possibilities. If the seventh digit is 3, then these are:

789654321; 987654321; 189654327; 981654327.

It can be easily verified that the seventh slice is not divisible by 7 in any of these cases.

If the seventh digit is 7, then the following numbers are obtained:

189654723; 981654723; 183654729; 381654729.

Here we find a solution: the seventh slice of the last number is also divisible by 7: $3816547=7 \cdot 545221$.

Therefore, there is only one number that satisfies the conditions of the problem, and that is 381654729.