Initial digit

Let the positive integer be $N = 6\times 10^m+n$ , where $m$ and $n$ are positive integers. Then we have:

$\begin{aligned} 6\times10^m + n - 6 \times 10^m & = \frac {6\times10^m + n}{25} \\ 25 n & = 6\times10^m + n \\ 24 n & = 6\times10^m \\ \implies n & = \frac {6\times10^m}{24} = \frac {10^m}4 \end{aligned}$

For $n$ to be an integer, the smallest $m=2$ and $n = \dfrac {100}4 = 25$ and $N = 6 \times 10^2 + 25 = \boxed{625}$ .

Let $N=10^{x} \cdot 6 + y$ , where $y<10^{x}$ . And then, we know that $y=\frac{1}{25}N$ $y=\frac{1}{25}(10^{x} \cdot 6 + y)$ $25y=10^{x} \cdot 6 + y$ $24y=10^{x} \cdot 6$ $4y=10^{x}$ $y=2^{x-2} \cdot 5^{x}$ $10^{x} \cdot 6 + y=10^{x} \cdot 6 + 2^{x-2} \cdot 5^{x}$ Then, we'll get minimal value of $y$ if $x$ is also minimal. That is, $x=2$ . And then we get $y=5^{2}=25$ and $y<10^{2}=100$

So, the number is $\boxed{625}$