 Solution

Again use integer coefficients after clearing denominators and put $S={\sum }_i{a}_i$. If $S=0$, choose any odd positive integer $u$ and take all $x_i=u$. This gives a monochromatic solution in any colouring.

For necessity, suppose $S\ne 0$ and choose $K$ with $2^K>|S|$. Colour the odd positive integers by their residue classes modulo $2^K$. In any monochromatic candidate, $x_i\equiv u(\mathrm{mod}{2}^K)$ for a common odd $u$. The equation would imply $uS\equiv 0(\mathrm{mod}{2}^K)$. Since $u$ is coprime to $2^K$, this forces $2^K\mid S$, impossible for nonzero $S$ with $|S|<2^K$. Thus $\boxed{\text{regular over odd positive integers}⟺\sum _i{a}_i=0}$. This is one-equation partition regularity over odd integers; its obstruction concerns the full coefficient sum rather than just a zero-sum subset.