Consider the cyclotomic field $K=\mathbb{Q}(\zeta_5)$.

**Proposition.** There is no $\alpha \in \mathcal{O}_K$ of degree 4 such that $D(\alpha)$ divides $N(\alpha)$.

**Proof.** Assume such an $\alpha$ exists. Since the discriminant of $K$ is $\Delta_K=5^3$, we must have $5^3|D(\alpha)$ and thus $5^3|N(\alpha)$. Let $\pi$ be the unique prime ideal above $5$ in $\mathcal{O}_K$, so that $5\mathcal{O}_K=\pi^4$. Then $\alpha \in \pi^3$. But we have surjective maps
\begin{equation*}
\frac{\mathcal{O}_K}{\mathbb{Z}[\alpha]} \to \frac{\mathcal{O}_K}{\mathbb{Z}+\alpha\mathcal{O}_K} \to \frac{\mathcal{O}_K}{\mathbb{Z}+\pi^3}.
\end{equation*}
The last group has cardinality $5^2$ since the image of $\mathbb{Z}$ in $\mathcal{O}_K/\pi^3$ is isomorphic to $\mathbb{Z}/5\mathbb{Z}$. We deduce that the index of $\mathbb{Z}[\alpha]$ in $\mathcal{O}_K$ is divisible by $5^2$. It follows that $5^4 \Delta_K | D(\alpha)$ and thus $5^7 | N(\alpha)$. Repeating the process, we get $N(\alpha)=0$, a contradiction.

I guess that if you take a quartic field $K$ with no intermediate subfield and in which some prime $p$ is sufficiently ramified, then the same reasoning will give you an example of number field for which the answer to your question is negative.