Find all positive integers that satisfy the following conditions: its first digit is 6, and removing this 6, the resulting integer is $\frac{1}{25}$ of the original integer;

Prove that there is no such integer: its first digit is 6, and removing its first digit results in $\frac{1}{35}$ of the original integer.
(1) Let the number sought be an $n$-digit number $m$, then $m=6 \times 10^{n-1}+\frac{m}{25}$. Thus, $m=\frac{25}{4} \times$ $10^{n-1}=625,6250,62500, \cdots$.
(2) The equation $m=6 \times 10^{n-1}+\frac{m}{35}$ is equivalent to $17 m=3 \times 10^{n-1}$, but $17 \nmid 3 \times 10^{n-1}$, so the equation has no solution.