List 4. (IMO-4-1) Find the smallest natural number $n$ whose decimal representation ends with 6, such that when the last digit 6 is deleted and written at the beginning of the remaining digits, it becomes four times $n$.
Solve for $n$ whose decimal representation ends with 6, meaning that it leaves a remainder of 6 when divided by 10. By the division algorithm, we have
$$
n=10 m+6 \text {. }
$$

where $m$ is a positive integer, and we assume it is an $l$-digit number. According to the problem,
$$
4(10 m+6)=6 \times 10^{l}+m .
$$

Thus,
$$
m=\frac{2\left(10^{2}-4\right)}{13} .
$$

For $m$ to be an integer, we need $13 \mid 2\left(10^{\imath}-4\right)$. Since 13 is a prime number, and $13 \times 2$, it follows that $13 \mid 10^{2}-4$ (here we use the property of prime numbers). Furthermore, to make $n$ the smallest, from the above equation, this is equivalent to making $m$ the smallest. Therefore, our problem is reduced to finding the smallest positive integer $l$ such that $13 \mid 10^{2}-1$.

By taking $l=1,2, \cdots$, and verifying one by one, we find that the smallest $l$ that satisfies the condition is $l=5$ (we used the "smallest" constraint). At this point, $m=15381$, and the smallest $n$ we seek is $153815$.

Comment: The numbers involved in this problem are relatively large, so verifying them is not difficult. This kind of "trial method" can be easily solved with the help of a computer for large number calculations. However, for certain theoretical problems, a computer is still powerless.

The following problem is closely related to Example 4 and is provided for readers to think about:

Problem. Prove: There exist infinitely many natural numbers $n$ that satisfy the conditions of the problem.