My progress on this so far:
($\mathbb{Z}$/$n\mathbb{Z}$, $+$) is basically the cyclic group $C_n$. So the question boils down to if there is an element $x$ of order $n$ for any $n$ in the group $(\mathbb{Z}$/$m\mathbb{Z})^\times $, because then $\langle x\rangle = (x, x^2, x^3,...,x^{m-1}, e)$ forms a cyclic group of order $m$.
The question can then also be formulated as the problem of finding $S$ where $$S = \{ n : \not\exists a, m \in \mathbb{Z}^+ s.t. a^n \equiv 1\ (mod\ m)\}$$
Or as
$$ S = \{ n : n\ \text{is not the order of any number a mod any number m }\}$$
From this I can definitely conclude that $x \in S\ if\ \forall n\ x \nmid \phi(n) $, because of Euler's Theorem, but I guess that doesn't really give us anything.
I'm not sure if this is the way I should be going about the problem, and if it is then I am not sure on how I should proceed further.
Outline of proof:
1) for prime $p > 2$ $(\Bbb Z/p^k)^*$ is cyclic (of order $p^{k-1}(p-1)$, obviously)
2) $(\Bbb Z/2^k)^*$ is $\Bbb Z/2$ times cyclic
(it's part of Gauss primitive root theorem)
3) use Chinese remainder theorem to obtain general description of $(\Bbb Z/n)^*$