Let DOID be an ordered integral domain with the property that there is no element strictly between any n*1 and the next (n+1)*1 multiples of 1. Please prove that DOID has the well-ordering property.
I know that the $\Bbb Z$, as well as the Q ordered rational numbers satisfy the properties of an ordered integral domain. Once we impose on DOID also a well ordering property, the extended properties characterize the integer numbers: Q no longer satisfies the extended axioms. The question is wether the above mentioned property (no element strictly between one multiple and its next) is enough to impose discreteness and thus stop Q to satisfy the extended axioms, in the same way that the well-ordering property does.
As William Elliot points out, $\mathbb{Z}$ itself is not well-ordered. Actually, the only well-ordered ring is the trivial one, for if an ordered ring contains a non zero element $a$, then either $a > 0$ and $\{-n.a \ | \ n \in \mathbb{N}\}$ has no minimum, or $a < 0$ and $\{n.a \ | \ n \in \mathbb{N}\}$ has no minimum.
As for your statement implying that DOID$_{\geq 0}$ is well-ordered, this fails as well. For instance take $\mathbb{Z}[X]$ ordered by $P \prec Q$ iff $P(n) < Q(n)$ for sufficiently large integer $n$. This ring satisfies your property but $\{X - n \ | \ n \in \mathbb{N}\} \subset \mathbb{Z}[X]_{\geq 0}$ has no minimum.
The rings in which your property works are those whose fraction field is archimedean: those where $\forall x > 0, \exists n \in \mathbb{N}_{>0}, 1 < n.x$ and $x < n.1$.