Let $\chi : (\mathbb Z/f\mathbb Z)^\times \to K = \mathbb Q(\mu_{\phi(f)})$ be a primitive Dirichlet character. Assume moreover that it is *not* quadratic, that is, $\chi^2$ is not the trivial character. Let $\pi_1,\dots,\pi_g$ be the primes lying over $2$ and $v_1,\dots,v_g$ be the corresponding valuations. Recall that:
$$L(0,\chi) = \frac1f\sum_{n=1}^fn\chi(n)$$

Experimentally (upto conductor 200), I find that there always exists some $k$ such that $v_k(L(0,\chi)) > 0$. Does anyone know a proof?

Note that it is not true that $2 | L(0,\chi)$. For instance, for for the character of conductor $5$ mapping $2 \to i$, we have $L(0,\chi) = (i+3)/5$. There are lots of other examples.

Note also that we do require the condition that $\chi$ is non quadratic. For instance, if $f = p \equiv 3 \pmod4$ and $\chi$ is quadratic, then: $$pL(0,\chi) \equiv (p-1)/2 \equiv 1 \pmod 2.$$

I asked this question a few hours before on stackexchange but at the suggestion of someone, I am posting it here. I have deleted the question on stackexchange.