= Solution
The <Ohnishi orderability criterion> states that, for each finite list $g_1,\ldots,g_n\ne1$, there are signs $\varepsilon_i\in\{-1,1\}$ such that the <semigroup> of nonempty products of conjugates of $g_i^{\varepsilon_i}$ does not contain $1$. The conjugates are essential for a two-sided invariant <total order>. Necessity follows by choosing all signed elements positive: their conjugates and nonempty products are still positive.
For sufficiency, consider the compact product $\{+,-\}^{G\setminus\{1\}}$. Impose the closed conditions that an inverse has the opposite sign, that conjugation preserves sign, and that a product of two positive elements is positive whenever it is nonidentity. A positive pair cannot multiply to the identity because of the inverse condition. Each condition uses finitely many coordinates. Given finitely many conditions, collect every nonidentity element appearing in them into a finite list and choose signs using the criterion. These signs satisfy the conditions: an incorrectly negative product would contribute its positive inverse to the generated <semigroup>, producing identity with the positive factors; an incorrect conjugate or inverse gives the same contradiction. Extend the signs arbitrarily to the remaining coordinates. Thus the finite intersection property and compactness give a global sign assignment. Its positive elements form a conjugation-invariant <semigroup> $S$ with $S\cap S^{-1}=\varnothing$ and $S\cup S^{-1}=G\setminus\{1\}$. Defining $a<b$ by $a^{-1}b\in S$ gives the required <linearly ordered group>.
Restriction shows that every <finitely generated subgroup> of an orderable <group> is orderable. Conversely, suppose every such subgroup is orderable. If the criterion failed on $g_1,\ldots,g_n$, each of the finitely many sign choices would have an identity witness made of finitely many conjugates. Collect all $g_i$ and all conjugating elements in these witnesses into one <finitely generated subgroup> $H$. An order on $H$ supplies one positive sign choice, but its corresponding witness is a nonempty product of positive elements in $H$, a contradiction. Hence the criterion holds in $G$.
Finally, a <finitely generated subgroup> of a <torsion-free abelian group> is $\mathbb Z^r$ by the <Fundamental theorem of finitely generated abelian groups>. Its <lexicographic order> is two-sided invariant. Applying the local-to-global conclusion proves that every <torsion-free abelian group> admits the required order.
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