= 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 $S\ne0$ and choose $K$ with $2^K>|S|$. Colour the odd positive integers by their <residue classes> modulo $2^K$. In any <monochromatic> candidate, $x_i\equiv u\pmod{2^K}$ for a common odd $u$. The equation would imply $uS\equiv0\pmod{2^K}$. Since $u$ is <coprime> to $2^K$, this forces $2^K\mid S$, impossible for nonzero $S$ with $|S|<2^K$. Thus \b[$\boxed{\text{regular over odd positive integers}\iff\sum_i a_i=0}$]. This is <one-equation partition regularity over odd integers>; its obstruction concerns the full coefficient sum rather than just a zero-sum subset.
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