$k$ with an even sum of digits for all multiples of $k$? Is there a number $k\in\mathbb{N}$ such that $k\cdot n$ has an even sum of digits for all $n\in\mathbb{N}$?

I would be grateful for any ideas of how to attack this problem...
There is no such $k \in \mathbb{N}$ : given any $k$, we can calculate a value $n$ such that $kn$ has an odd sum of digits.

Factorize $k$ as $k = 2^a 5^b m$. Then consider the following integer

$$q = 10^{\max(a, b)} \sum_{i=1}^{m} {\left(10^{\phi(m)}\right)}^{i}$$

where $\phi(m)$ is Euler's totient function. We claim that $q$ is divisible by $k$. Since $\gcd(2^a 5^b, m) = 1$, if we can show that $2^a 5^b \mid q$ and $m \mid q$ we will be done.

The left term in the product is divisible by $2^a 5^b$ by construction; now we consider the right term.

Since $\gcd(m, 10) = 1$, Euler's theorem tells us that $10^{\phi(m)} \equiv 1 \pmod m$. Additionally, for any power of $i$, $1^i \equiv 1 \pmod m$. So all $m$ summands have remainder one, meaning that the right term is equivalent to $1 \cdot m \equiv 0 \pmod m$, and $m$ divides $q$.

Consequently, $q$ is divisible by $k$, which we can write as $q = kn$ for some integer $n$.

What is the sum of the digits in $q$? The right term is a sum of $m$ distinct powers of ten, whose digits will sum to $m$. Multiplying by the left term, a power of ten, doesn't change the sum, and so the sum remains $m$ – which is odd by definition.
There is no such number. All the number crunching in the world is no proof, but when someone crunches up to a largish number, it looks very unlikely.

The way I would start to attack this problem is to show that for every positive $k$ there are infinitely many $kn$ with odd digital root. It's of course possible for $kn$ to have an odd digital root but an even digit sum (e.g., $99$).