Given are $n$ distinct natural numbers. For any two of them, the one is obtained from the other by permuting its digits (zero cannot be put in the first place). Find the largest $n$ such that it is possible all these numbers to be divisible by the smallest of them?
1. Let the smallest number among the \( n \) distinct natural numbers be \( x \).
2. Since all numbers are permutations of \( x \), they must have the same digits in some order. Therefore, they must have the same number of digits.
3. For any two numbers \( a \) and \( b \) in this set, \( a \) is obtained by permuting the digits of \( b \). Hence, \( a \equiv b \pmod{x} \).
4. This implies that \( x \) must divide any permutation of its digits. 

To find the largest \( n \) such that all \( n \) numbers are divisible by \( x \), we need to consider the constraints on the number of digits and the properties of the numbers.

5. Consider the case \( n = 9 \):
   - If \( n = 9 \), then there are 9 distinct permutations of \( x \).
   - For \( x \) to divide all its permutations, \( x \) must be a number such that all its permutations are multiples of \( x \).
   - However, unless all the digits of \( x \) are the same (e.g., all 1's), \( 9x \) will have more digits than \( x \), which is contradictory.
   - Additionally, a number like \( 111111111 \) (9 ones) does not meet the requirement because its permutations are not distinct natural numbers.

6. Consider the case \( n = 8 \):
   - For \( n = 8 \), we need to find a number \( x \) such that there are 8 distinct permutations of \( x \) and all are divisible by \( x \).
   - One such number is \( x = 142857 \), which is obtained from the decimal expansion of \( \frac{1}{7} \). The number \( 142857 \) has the property that its cyclic permutations are multiples of \( 142857 \):
     \[
     142857 \times 1 = 142857
     \]
     \[
     142857 \times 2 = 285714
     \]
     \[
     142857 \times 3 = 428571
     \]
     \[
     142857 \times 4 = 571428
     \]
     \[
     142857 \times 5 = 714285
     \]
     \[
     142857 \times 6 = 857142
     \]
     \[
     142857 \times 7 = 999999 \quad (\text{not a valid permutation})
     \]
   - However, the number \( 142857 \) does not provide 8 distinct permutations that are all multiples of \( 142857 \).

7. Consider the number \( x = 1176470588235294 \), which is obtained from the decimal expansion of \( \frac{2}{17} \):
   - The number \( 1176470588235294 \) has the property that its cyclic permutations are multiples of \( 1176470588235294 \):
     \[
     1176470588235294 \times 1 = 1176470588235294
     \]
     \[
     1176470588235294 \times 2 = 2352941176470588
     \]
     \[
     1176470588235294 \times 3 = 3529411764705882
     \]
     \[
     1176470588235294 \times 4 = 4705882352941176
     \]
     \[
     1176470588235294 \times 5 = 5882352941176470
     \]
     \[
     1176470588235294 \times 6 = 7058823529411764
     \]
     \[
     1176470588235294 \times 7 = 8235294117647058
     \]
     \[
     1176470588235294 \times 8 = 9411764705882352
     \]
   - These are 8 distinct permutations of \( 1176470588235294 \) that are all multiples of \( 1176470588235294 \).

Therefore, the largest \( n \) such that it is possible for all these numbers to be divisible by the smallest of them is \( n = 8 \).

The final answer is \( \boxed{8} \).