4. 252 (1) Find all positive integers whose first digit is 6, and when this 6 is removed, the resulting integer is $\frac{1}{25}$ of the original integer.
(2) Prove that there is no such positive integer whose first digit is 6, and when this 6 is removed, the resulting integer is $\frac{1}{35}$ of the original integer.
［Solution］ Let the positive integer satisfying the given conditions be
$6 \cdot 10^{n}+m$, where $n \in N$ and $0 \leqslant m \leqslant 10^{n}-1$.
(1) From the given conditions, we have
$$m=\frac{1}{25}\left(6 \cdot 10^{n}+m\right)$$

Simplifying, we get
$$m=2^{n-2} \cdot 5^{n}$$

Therefore, the number we are looking for is
$$6 \cdot 10^{n}+2^{n-2} \cdot 5^{n}=625 \cdot 10^{n-2}$$

That is, $625, 6250, 62500, \cdots$
(2) From the given conditions, we have
$$m=\frac{1}{35}\left(6 \cdot 10^{n}+m\right)$$

Simplifying, we get $17 m=3 \cdot 10^{n}$.
The left side of the equation is a multiple of 17, while the right side cannot be divided by 17, which is a contradiction. Therefore, there does not exist a positive integer that satisfies the given conditions, and the proposition is proved.