$k$ such that $n,2n,\dots,kn$ have odd sum of digits is the following statement true or not?

for any $k\in\mathbb{N}$ there exists $n\in\mathbb{N}$ such that all numbers $n,2n,\dots,kn$ have odd sum of digits?

I have no idea... it may turn out hard (for even sum of digits it would be very easy)
Hint : if we consider the number $n=9\cdots9$ with $2k-1$ of $9$'s. so for every element $0\leq i\leq 10^k$ we have the sum of digits of $ik$ is odd (is equal to 9(2k-1)) which proves your statement. I will provide a proof if this is necessary.
Hint: Having an odd sum of digits means is equivalent to the condition that the number mod 9 is odd. This means that it's sufficient to consider $n \in \{0,\ldots,8\}$.