Let $q$ and $r$ be distinct prime numbers. I noticed (computing a few cases) that $\zeta_{2q} + \zeta_{2q}^{1} + \zeta_{2r} + \zeta_{2r}^{1}$ is a unit (in $\mathbb{Z}[\zeta_{2qr}]$, say). Is this always true? Why is that?

I assume you want $q$ and $r$ to be odd primes. Also, note that I will be using the notation that $\zeta_m$ means an arbitrary primitive $m$th root of unity (but the same one every time it appears in an equation), and will be proving the statement in that generality. Lemma: For any odd $m>1$ and any $\zeta_m$, the number $\zeta_m+1$ is a unit. Proof: Let $r$ be such that $m  2^r1$. We'll abbreviate $\zeta_m$ to $\zeta$. Then $\zeta^{2^r} = \zeta$ so $$1 = \left( \frac{\zeta^{2}1}{\zeta 1} \right) \left( \frac{\zeta^{4}1}{\zeta^{2} 1} \right) \cdots \left( \frac{\zeta^{2^r}1}{\zeta^{2^{r1}} 1} \right)=$$ $$\left( \zeta+1 \right) \left( \zeta^{2} + 1 \right) \cdots \left( \zeta^{2^{r1}} +1 \right),$$ exhibiting an explicit inverse for $\zeta+1$. Let $\eta$ be a primitive $2qr$ root of unity. Then your proposed unit is $\eta^{r}+\eta^{r} + \eta^q + \eta^{q}$ and factors as $$\eta^r (1+\eta^{qr})(1+\eta^{qr}).$$ Since $q$ and $r$ are odd and relatively prime, $\eta^{qr}$ and $\eta^{q+r}$ are primitive $qr$th roots of unity and we are done by the lemma. 


The answer is also yes if one of the primes, say $r$, is $2$, because then $\zeta_{2r}+\zeta_{2r}^{1}=0$ and $\zeta_{2q}+\zeta_{2q}^{1}=\zeta_{2q}(1+\zeta_{2q}^{2})$ is a unit (as $\zeta_{2q}^{2}$ is a primitive $q$th root of $1$ and $q$ is an odd prime). Edit: (1) Note that if both primes are odd then $\zeta_q+\zeta_q^{1}+\zeta_r+\zeta_r^{1}$ is also a unit. Indeed, $\zeta_q$ is a primitive $2q$th root of $1$ (this relies on $q$ being odd), so let's call it $\zeta_{2q}$, and likewise for $r$. Then $(\zeta_q+\zeta_q^{1}+\zeta_r+\zeta_r^{1})=\zeta_{2q}+\zeta_{2q}^{1}+\zeta_{2r}+\zeta_{2r}^{1}$, and we know that the latter is a unit. (2) Note also that if one of the primes, say $r$, is $2$ then $\zeta_q+\zeta_q^{1}+\zeta_r+\zeta_r^{1}$ is not a unit: it is equal to $\zeta_q+\zeta_q^{1}2$, so it is a unit times the square of $\zeta_q1$, but the latter (unlike $\zeta_q +1$) is not a unit because it goes to $0$ under the unique ring homomorphism $\mathbb Z [\zeta_q]\to \mathbb Z/q$, which takes $\zeta_q$ to $1$. 

