7. Given $n$ distinct natural numbers, any two of which can be obtained from each other by permuting the digits (zero cannot be placed in the first position). For what largest $n$ can all these numbers be divisible by the smallest of them?
Answer: 9. It is clear that there cannot be more than nine numbers.

We will use a known property of the period of a purely periodic rational fraction $\alpha=$ $a / b<1$ with coprime $(a, b)$: the length of the period is the smallest natural $t$ for which $\left(10^{t}-1\right) \vdots b$, and the period $T$ itself is the number $\left(10^{t}-1\right) \alpha$.

From this, it follows that if $a_{1} / b$ and $a_{2} / b$ are two proper irreducible fractions, and $a_{1}=k a_{2}$, then the period of the first fraction is exactly $k$ times the period of the second.

Consider the fractions 2/19, 4/19, 6/19, 8/19, 10/19, 12/19, 14/19, 16/19, 18/19. They have periods of the same length, and all of them are divisible by the period of 2/19, which is 105263157894736842. On the other hand, it can be verified that 10 is a primitive root modulo 19. Therefore, the numbers $2 \cdot 10^{s} / 19$ for $0 \leqslant s \leqslant 17$

It can be verified that 10 is a primitive root modulo 19. Therefore, the numbers $10^{s} / 19$ for $0 \leqslant s \leqslant 17$ after removing the integer part are all 18 proper irreducible fractions with the denominator 19. Their periods have the same length and are obtained from the period of the fraction $2 / 19$ by cyclic shifts of $s$ digits to the right. In particular, they all consist of the same set of digits. (And all, except $1 / 19$, do not start with zero, i.e., they represent natural numbers of the same length). From the previous paragraph, it follows that the periods of the fractions $4 / 19,6 / 19,8 / 19,10 / 19,12 / 19,14 / 19,16 / 19,18 / 19$ are divisible by the period of the fraction $2 / 19$ and together they form an example of 9 numbers.