Hope that my answer is correct. Ans $=4$ Assume $D(n)$ denote the leftmost digit of $n$. Consider $D(n)=5, 6, 7, 8, 9$, then the number of different digits of $D(n), D(2n), ..., D(9n)$ must be larger than $4$. This is because $D(2n)=1$, $D(9n)=D(n)-1$, so all numbers from $1$ to $D(n)-1$ will appear at least once. We will prove that there's no number with the number of different leftmost digits less than $4$. If $n$ starts with $10$ to $32$, then $D(n)\ne D(2n)\ne D(3n)\ne D(4n)$ But if $n$ starts with $33$ to $49$, then $D(n)\ne D(2n)\ne D(3n)\ne D(9n)$ So the minimal number is $4$. Consider $n=25$, then we have $25, 50, 75, 100, 125, 150, 175, 200, 225$