$4 \cdot 252$ (1) Find all positive integers, the first digit of which is 6, and removing this 6, the resulting integer is $\frac{1}{25}$ of the original integer.
(2) Prove: There is no such positive integer, the first digit of which is 6, and removing this 6, 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
$$
m=\frac{1}{25}\left(6 \cdot 10^{n}+m\right) \text {, }
$$

which simplifies to
$$
m=2^{n-2} \cdot 5^{n} .
$$

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

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

which simplifies to $17 \mathrm{~m}=3 \cdot 10^{n}$.
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 satisfying the given conditions, and the proposition is proved.