Let $A$ be a simple linearly ordered complete divisible archimedean abelian group.
It is in particular a $\mathbb Q$-vector space. For any $a\in A$, we have an embedding $\mathbb Qa\to A$ which automatically extends to $\mathbb R\cong \mathbb Ra\to A$, as $A$ is complete.
I claim that this map must be an isomorphism. Clearly it must be injective as $\mathbb R$ is simple. Moreover, if $b\in A$, $b\leq na$ for some $n$, and so we can actually consider $\{r\in \mathbb R, ra < b\}$ and this has an upper bound, so it has a least upper bound, $r_b$.
Now for any $r< r_b$ we have $(r_b-r)a > r_ba - b$, so for any $n\in \mathbb N$, $a> n(r_ba-b)$. By the archimedean character of $A$, it follows that $r_ba - b \leq 0$, i.e. $r_ba \leq b$ as well.
But now one can also similarly prove $r_ba \geq b$, and so $r_ba = b$
(a simpler way to phrase surjectivity would be to use Dedekind reals, but let me stick to this formulation in case you don't know these).
The claim follows, and so by embedding your linearly ordered group into its rationalization and then the completion of that, you get that $\mathbb R$ is weakly terminal among simple linearly ordered abelian groups.
Best Answer
Yes. If $x$ is positive then so is $nx$ for any $n \in \mathbb{Z}^+$ and if $x$ is negative then $-x$ is positive and $nx = 0$ iff $n(-x) = 0$.
In other words, a totally ordered abelian group is necessarily torsionfree. More interestingly, the converse also holds: any torsionfree abelian group can be totally ordered (in at least one way). See Section 17.2 of these notes for the proof.