= Solution
Clear a common denominator first, so that all coefficients are nonzero integers. This changes neither the solutions nor <partition regularity>. We prove the one-equation criterion directly; no form of <Rado's theorem> is used as an assumption.
For necessity, choose a <prime number> $p>\sum_i|a_i|$. Colour a positive integer by its first nonzero base-$p$ digit: writing $x=p^{v_p(x)}u$ with $p\nmid u$, use the colour $u\pmod p\in\{1,\ldots,p-1\}$. Suppose the equation has a <monochromatic> solution. Let $v$ be the smallest <P-adic valuation> of its entries and put $I=\{i:v_p(x_i)=v\}$, a nonempty set. Divide the equation by $p^v$ and reduce modulo $p$. All entries indexed by $I$ contribute the same nonzero digit $u$, and all other terms vanish. Hence
$$
u\sum_{i\in I}a_i\equiv0\pmod p.
$$
It follows that $p$ divides $\sum_{i\in I}a_i$. The absolute value of this sum is less than $p$, so \b[$\boxed{\sum_{i\in I}a_i=0}$].
For sufficiency, take a nonempty zero-sum index set $I$ and put $S=\sum_{i\notin I}a_i$. If $S=0$, the full coefficient sum is zero and the constant vector $(1,\ldots,1)$ is already a <monochromatic> solution in every colouring. Otherwise choose $j\in I$, put $t=|a_j|$, and set $b_j=-\operatorname{sgn}(a_j)S$, $b_i=0$ for $i\in I\setminus\{j\}$. Then
$$
\sum_{i\in I}a_ib_i=-tS.
$$
Add the same sufficiently large nonnegative integer $H$ to all the $b_i$, obtaining $c_i=b_i+H\geq0$. The sum is unchanged because $\sum_{i\in I}a_i=0$. Let $K=\max_{i\in I}c_i$.
Apply the scaled <Brauer progression theorem>, proved above from the permitted <Van der Waerden theorem>, to obtain one-colour numbers $td$ and $a,a+d,\ldots,a+Kd$. Define
$$
x_i=a+c_i d\quad(i\in I),\qquad x_i=td\quad(i\notin I).
$$
All entries are positive and <monochromatic>, and
$$
\sum_i a_ix_i=a\sum_{i\in I}a_i+d\sum_{i\in I}a_ic_i+tdS=0-tSd+tSd=0.
$$
Therefore \b[the row is <partition regular> exactly when some nonempty coefficient subset sums to zero]. Repetition of entries is allowed by the definition of <partition regularity>.
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