Q:

Why do we choose the standard total order on the integers?

I understand why the set of natural numbers $\mathbb N = \{ 0, 1, 2, \cdots \}$ is equipped with a total order. Indeed, every monoid has a pre-order, where $$n' \succeq n \quad \mathrm{if~and~only~if} \quad n' = n + m \quad \mathrm{~for~some~} m.$$ In the case of $\mathbb N$, this pre-order is a total order. 
However, the same construction does not result in a total order on the set of integers $\mathbb Z$. Indeed, this set is a group, so its canonical monoid pre-order is trivial. i.e., $n' \succeq n$ for all $n, n' \in \mathbb Z$, since $n' = n + (n' - n)$.
Nonetheless, $\mathbb N$ is a subset of $\mathbb Z$, so it makes sense to assign an order relation to $\mathbb Z$ which extends the natural order on $\mathbb N$. I see (at least) two natural ways to do this:

The standard total order $\ge$ on $\mathbb Z$.
The pre-order on $\mathbb Z$ which corresponds to the absolute value norm. i.e., $n' \succeq n$ if and only if $|n'| \succeq |n|$. 

There is obvious pragmatic justification for choosing the standard total order; it's utility is not in question. However, there are also pragmatic advantages for the alternate pre-order. For example, it admits $0$ as a minimal element ($n \succeq 0$ for all $n \in \mathbb Z$), and it extends the canonical pre-orders on the monoids $\mathbb N$ and $-\mathbb N$. It also generalizes nicely to higher-dimensional settings such as $\mathbb Z^d$, where no natural total order exists. 
The integers exist in a universal mathematical sense: they form the Grothendieck group for the natural numbers. However, there seem to be (at least) two order-theoretic models for the integers: the totally ordered set $(\mathbb Z, \ge)$ and the pre-ordered set $(\mathbb Z, \succeq)$. 
I taught an undergraduate discrete mathematics course last semester and the book never even acknowledged the second model, nor did it provide justification for the first. I suppose this is acceptable for undergraduates, but as a mathematician I am bothered by the implicit choice for an ubiquitous mathematical structure.
Ergo my question:

What is a mathematical justification for "always" choosing the "standard order" $(\mathbb Z, \ge)$? 

That is, is there a universal characterization of this order structure which can be adapted to the setting of a Grothendieck group over general monoid? When does it it result in a total or partial order on the group instead of just a pre-order?

A:

The standard order is (up to isomorphism) the only total order on $\mathbb Z$ that makes it an ordered group under addition.  So I'd expect it to be useful in situations where addition plays a role; these are probably most (though certainly not all, as Ryan Budney pointed out in a comment to the question) of the situations that arise in practice

A:

Maybe the following can be an  justification:
A linear order of a semigroup can be continued by one and only one way to a linear order of its semigroup of fractions (L. Fuchs, Partially Ordered Algebraic Systems, Theorem III.X.4).
Note. Here the semigroup can be neither commutative nor cancellative.

A:

I assume (your) monoids are cancellative. 
Then the pre-order you define is a (partial) order if and only of the monoid $M$ is reduced (i.e. has no invertible elements besides the neutral one). 
For getting an order on the Grothendieck group $G$ say the element in $M$ are "positive elements" and define on $G$ the relation $g\ge h$ if $g-h \in M$. This extends the preorder of $M$; is transitive and reflexive; and it is anti-symetric if and only if $M$ contains no non-trivial invertible elements.   
It is a total order if $g$ or $-g$ in $M$ for each $g \in G$; such a monoid $M$ is sometimes called a valuation monoid (in analogy with valuation rings, which have this property with respect to their quotient field, for multiplication of course; thus also these monoids are more frequently noted multiplicatively). 

