A group $G$ admits a nontrivial preorder iff it admits a nontrivial order-preserving action on a totally ordered set (which can be chosen to be $\mathbf{Q}$ or its completion $\mathbf{R}$ if $G$ is countable).

Proof:
[I like left actions rather than right actions so I'll go ahead with left-invariant instead of right-invariant: you can pass from one to another by inversion.]

Indeed suppose that $G$ admits such an action. Let $x$ be not fixed by all of $G$. Define $g\le h$ if $gx\le hx$. Then this is a non-trivial left-invariant total preorder on $G$.

Conversely, suppose that $G$ has a left-invariant total preorder $\le$. Let $H$ be the set of elements $h$ such that $1\le h\le 1$. Then $H$ is a subgroup: left-multiplying the latter by $h^{-1}$ shows it's stable by inversion, and if $1\le g\le 1$ as well, multiplying the former on the left by $g$ yields $g\le gh\le g$, whence $1\le gh\le 1$.

The preorder is actually right-$H^2$-invariant, in the sense that $g\le g'$ and $h,h'\in H$ implies $gh\le g'h'$. Indeed, we have $gh\le g\le g'\le g'h'$.
The coset space $G/H$ inherits a $G$-invariant total order by $gH\le g'H$ iff $g\le g'$, this does not depend on the choices of $g,g'$ because of the above right $H^2$-invariance.

If moreover $G/H$ is countable, you can just take the lexicographic product $(G/H)\times\mathbf{Q}$ (with trivial action on $\mathbf{Q}$) to get an action on $\mathbf{Q}$. Taking completion also yields an action on $\mathbf{R}$.

In some cases you have an even simpler characterization. Indeed, a theorem of Witte (*Alg. Geom. Topol.* 2006, arXiv link; ProjectEuclid) can be restated as: a *finitely generated amenable* group $G$ admits a nontrivial order-preserving action on some totally ordered set (or on the real line, it's the same) if and only if it admits $\mathbf{Z}$ as a quotient.

Edit: Here's a direct proof (not using actions) that if $G$ admits a nontrivial left-invariant total preorder $\le$, then it admits a nontrivial left-orderable quotient: fix a strict well-ordering $\prec$ on the set $G$ (unrelated to the group structure and $\le$: if $G$ is countable just take an enumeration of $G$), and say that $g\not\le' g'$ if there exists $h\in G$ such that $gh\not\le g'h$ and $gh'\le g'h'\le gh'$ for every $h'\prec h$. Then $\le'$ is another left-invariant total preorder, but has the additional feature that the set $N$ of $g$ such that $1\le' g\le' 1$ is a normal subgroup ($N$ is indeed, using that $\prec$ is a well-ordering, the set of $g$ such that $1\le h^{-1}gh\le 1$ for all $h\in G$). Moreover $N$ is contained in the subgroup $\{g:1\le g\le 1\}$ and hence is not all of $G$. So the quotient $G/N$ is a nontrivial left-orderable group.

linearpre-orders, and also sometimes they are calledtotalpre-orders. $\endgroup$faithfulorder-preserving action on a totally ordered set and to be left-orderable are equivalent. So it follows from my answer that your condition (being totally left-preorderable) is indeed equivalent to the existence of a nontrivial left-orderable quotient. $\endgroup$