We have, for $z = e^{\frac{2 \pi i}{17}}$, $$(z + z^2 + z^4 + z^8 + z^{16} + z^{15} + z^{13} + z^9) (z^3 + z^5 + z^6 + z^7 + z^{10} + z^{11} + z^{12} + z^{14}) \in \mathbb Z.$$ This happens because the Galois group $\operatorname{Gal}(\mathbb Q(z)/\mathbb Q)$ acts on the partition of roots by either swapping them or invariant. For the general case with primitive roots, it suffices to find a quotient group of $\operatorname{Gal}(\mathbb Q(z)/\mathbb Q) \cong (\mathbb Z/n\mathbb Z)^\times \cong (\mathbb Z/\varphi(n) \mathbb Z, +)$ isomorphic to $\mathbb Z/ 2 \mathbb Z$. Since the Euler totient is always even, this is achievable.
What happens if we require all non-trivial roots of unity (i.e. not equal to 1)? This would require a partition such that the group action of $(\mathbb Z/n\mathbb Z)^\times$ on $\{1,2,\dots, n-1\}$ either swaps the sets or is invariant. Or in more abstract terms, the $G$-set has a quotient with exactly two elements. There is always a boring solution: put primitive roots in one group, and non-primitive ones in the other. This corresponds to the trivial quotient $G$-set.
If we require the partition to be non-trivial (i.e. the group acts non-trivially on the partition, i.e. each factor must not be integers themselves), this doesn't always happen, for example when $n = 15$. (Note that the roots of unity are algebraic integers, so we just need to guarantee rationality.) So the interesting part of the question reduces to
When can non-trivial roots of unity be partitioned into two parts, such that the sum of each part is not an integer, but their product is?
Or equivalently
Let $G = (\mathbb Z/ n \mathbb Z)^\times$. When does the $G$-set $\{1,\dots,(n-1)\}$ (whose action is given by modular multiplication) have a quotient with two elements but not isomorphic to $1+1$?
I found this problem when investigating a typo from a more elementary question. I should be able to figure this out tomorrow, and I'll self-answer then.
The only such numbers are odd prime powers.
Let me first reduce the case to elementary number theory. The orbits of the $G$-set are various quotients of $G$, and we need to pick another quotient isomorphic to $\mathbb Z/2\mathbb Z$ that lies below all those quotients. Dually, we have a lot of subgroups of $G$, and we wish to select an index-2 subgroup containing all of them.
These subgroups are precisely $G_k = \{mk+1 \pmod n \mid m \in \mathbb Z\} \cap (\mathbb Z/n\mathbb Z)^\times$, where $k$ runs through the non-trivial factors of $n$. It is now easy to see that if $n$ has a lot of factors, then $\bigcup_k G_k$ will inevitably generate the entire group. For $n = p^k$, we have $\bigcup_{j=1}^k G_{p^j} = G_{p}$. This is a subgroup of index $(p-1)$, which is of course contained in a subgroup of index $2$. (Even numbers $n$ are out of the question.)
Suppose $n$ is not a prime power, say $n = p^k m$ with $p \nmid m$. Then $$(ap^k + 1)(bm+1) \equiv (ap^k + bm) + 1 \pmod n,$$ which by Bézout's theorem covers all of $(\mathbb Z/n\mathbb Z)^\times$. We don't neeed to worry about our previous intersection with $(\mathbb Z/n\mathbb Z)^\times$, because if a product is coprime with $n$, then each factor must already be coprime, and hence the two factors will safely land in $G_{p^k}$ and $G_m$, respectively. Hence we have shown that non-prime factors will never do.