Is there a natural example of a divisible torsioned (= periodic) abelian group? The following are easy to verify:
(I) Every ordered group is torsion-free.
(II) Every non-trivial torsion-free group is infinite.  
Also, if $G$ is a non-trivial divisible group of a finite order $n,$ then for a non-zero element $a\in G$ we can not find an element $b$ with $a=\underbrace{b+\ldots+b}_{n-times}.$ Because, using Lagrange's theorem, the element $\underbrace{b+\ldots+b}_{n-times}$ would be equal to zero for any $b\in G.$ Hence, we also have the following:  
(III) Every non-trivial divisible group is infinite. 
Using a compactness argument from model theory and the fact that every finitely generated torsion-free abelian group is isomorphic to some $\mathbb{Z}^{n},$ we also have that:  
(IV) Every torsion-free abelian group is orderable. 
Hence, in the realm of abelian groups torsion-free coincides with orderable.
The following are some typical examples for the aforementioned notions in the realm of abelian groups:
1- $\langle\mathbb{Z},+,0\rangle$ is non-didivisible torsion-free.
2- $\langle\mathbb{Q},+,0\rangle, \langle\mathbb{R},+,0\rangle, $ and $\langle\mathbb{C},+,0\rangle$ all are divisible torsion-free.
3- For any $n,$ the group $\langle(\mathbb{Z}_{n})^{\omega},+,0\rangle$ is infinite non-divisible and torsion (=periodic).
The situation remains almost unchanged if you consider the above examples with their natural multiplications:
1- $\langle\mathbb{Z}^{+},.,1\rangle$ is not a group at all!
2- $\langle\mathbb{Q}^{+},.,1\rangle$ and $\langle\mathbb{R}^{+},.,1\rangle$ are torsion-free but non-divisible: there is no element like $a$ with $-1=a.a.$
3- $\langle\mathbb{C}^{+},.,1\rangle$ is torsion-free and divisible since, as a consequence of being an algebraically closed field, for every $a,$ the equation $x^n=a$ has always a solution in $\mathbb{C}^{+}.$
4- $\langle\mathbb{Z}^{+}_{n},.,1\rangle$ is a group if and only if $n$ is prime. In this case also, $\langle(\mathbb{Z}^{+}_{p})^{\omega},.,1\rangle$ is torsioned non-divisible.
Now, my question is if there exists a divisible torsioned(=periodic) abelian group built using numerical examples? In fact, due to some decidability concerns, I am more interested in additive examples.
 A: The simplest example is the quotient $\mathbb{Q}/\mathbb{Z}$, i.e. the additive rational numbers modulo $1$.  This is divisible for the same reason that $\mathbb{Q}$ is, and is torsion since any rational number has an integer multiple.
A: Jim's nice example can be easily generalized to any divisible torsion-free abelian group $G$ which essentially derived from the fact that any finitely-generated torsion-free 
abelian group is isomorphic to $\mathbb{Z}^{n}$ for some $n.$ More conceretely:
Let $G$ be a divisible torsion-free abelian group and $a\in G$ non-zero. Let $H$ be the subgroup generated by $a,$ namely:
$$H:=\Big\{ka:k\in\mathbb{Z}\Big\}.$$
If there are $m,n$ with their maximum strictly greater than 1 such that $\frac{a}{n}=ma,$ then we have that $(nm-1)a=0$ contradicting the fact that $G$ is torsion-free. So, the n-divisor of $a,$ i.e. $\frac{a}{n},$ does not belong to $H$.
On the other hand, it directly follows that the quotient group $G/H$ (like $\mathbb{Q}/\mathbb{Z}$ in Jim's answer) is divisible. But, we also have:
$$\underbrace{(\frac{a}{n}+H)+\cdots+(\frac{a}{n}+H)}_{n-times}=H.$$
Which according to the above discussion shows that $G/H$ is torsioned.
In fact, when $G=\langle\mathbb{Q},+\rangle,$ each of the 1-generated subgroups of $\mathbb{Q}$ can play the role of $H$; in particular all $n\mathbb{Z}$s.
