$\mathbb{Z}$ itself is Archimedean, but it is not divisible: for example, there is no $b \in \mathbb{Z}$ such that $2b = 1$.
The ring $Q[x]$, with the total ordering in which a polynomial in $x$ with rational coefficients is positive if and only if its trailing coefficient is positive, is divisible, but it is not Archimedean: for example, $0 < nx < 1$ for every positive integer $n$, because $1 - nx > 0$.
Further to my comment above, in response to the revised question,
these extracts are from F. Loonstra, "Ordered groups",
Proc. Nederl. Akad. Wetensch. 49 (1946), pp. 41-46 (PDF
here):
If an ordered set $O$ is dense in itself, then generally $O$ need
not be dense-ordered. (A set $O$ is called dense-ordered if for
each pair of elements $a$ and $b$ of $O$ there is at least one
element $c$ "between" $a$ and $b$.) For ordered groups however we
prove:
If $G$ is an ordered group dense in itself, then $G$ is densely
ordered.
If there was namely between two elements $a$ and $b$ no further
element, then every element had an immediate successor and
predecessor and thus $G$ was not dense in itself. Conclusion: An
ordered group $G$ is dense in itself if and only if $G$ is densely
ordered. If on the contrary one element is isolated, then every
element is necessarily isolated; consequently:
An ordered group $G$ is either a discretely ordered group, which
means that every element has an immediate successor and predecessor,
or a densely ordered group. The property of an ordered group as
being discrete resp. dense is as a topological property invariant
for a mapping preserving the order.
If we divide an ordered set $N$ into two subsets $N_1$ and $N_2$
with the conditions:
I. Every element of $N$ belongs to one, but only one of the sets
$N_1$ and $N_2$;
II. Neither $N_1$ nor $N_2$ is empty;
III. All elements of $N_1$ precede those of $N_2$.
then we call this division of $N$ a Dedekind-cut $N_1 | N_2$ in $N$.
There are only the following cases that exclude each other:
$1$. $N_1$ has a last, $N_2$ a first element; then we call the
cut $N_1 | N_2$ a sault in $N$.
$2^a$. $N_1$ has a last, $N_2$ no first element;
$2^b$. $N_1$ has no last, $N_2$ has a first element; we call
the cut in $2^a$ and $2^b$ a continuous cut.
$3$. $N_1$ has no last. $N_2$ has no first element; the cut is
called a lacuna in the set $N$.
The ordered set $N$ is called continuously ordered, if no cut in $N$
defines a sault or a lacuna.
A non-archimedean ordered group is not continuously ordered.
If the ordering was continuous then of every cut $A | B$ either $A$
had a last, $B$ no first element, or just the reverse. We shall try
to prove, however, that in every non-archimedean ordered group there
is to define a non-continuous cut. Suppose $e < a < b$ and $a$
non-archimedean with regard to $b$. Let the class $A$ contain all
elements $\leqslant a^n$ for a positive integer $n$, $B$ the
remaining elements; $B$ is not empty, as $b \in B$. $A$ has no last
element, nor has $B$ a first element, for $B$ also contains the
elements $ba^{-1} > ba^{-2} > \cdots$ elements that precede $b$.
Therefore a non-archimedean ordered group is never a continuous
ordered set or directly: continuously ordered groups are necessarily
archimedean ordered groups.
A densely ordered group $G$ has the property that to every element
$a > e$ there exists an element $b > e$, so that $b^2 < a$.
[A proof of that is provided, but ...]
Without a proof we pronounce the following theorem:
In a continuous ordered group there exists to every element $a > e$
one element $b > e$ with $b^n = a$.
[I was just typing this out, making edits and corrections, and joining
it all up, and didn't notice that he hadn't provided a proof of this part!
I think I have a proof in my own handwritten notes from years ago, but
it might not be easy to find, and may be wrong. Let's just hope the result
is obvious ... er ... an exercise for the reader?] :)
Noah Schweber's answer to the following question may help to fill in the, er, gap
I've just left:
Is $\mathbb{R}$ the only complete ordered Abelian group?.
