= Solution
Use the standard definition of a <partition regular matrix>: for every <finite coloring> of the <positive integers>, there is a positive solution whose coordinates all have one color. Coordinates may repeat. Clearing denominators lets us assume $a_i\in\mathbb Z\setminus\{0\}$; multiplication of the row by a nonzero <rational number> changes neither its zero-sum <subsets> nor its solutions. We prove both directions of the <Rado theorem for one equation> directly, without invoking any form of <Rado's theorem>.
For necessity, choose a <prime number> $p>\sum_i|a_i|$ and use the <last nonzero digit coloring>: write $x=p^{v_p(x)}u$, with $p\nmid u$, and assign color $u\bmod p$. Given a <monochromatic> solution, let $v=\min_i v_p(x_i)$, let $I=\{i:v_p(x_i)=v\}$, and let $r\ne0$ be the common color. Divide $\sum_i a_i x_i=0$ by $p^v$ and reduce modulo $p$. Terms outside $I$ vanish; the others give
$$
r\sum_{i\in I}a_i\equiv0\pmod p.
$$
The nonzero residue $r$ is invertible modulo the <prime number> $p$, so $p$ divides $\sum_{i\in I}a_i$. This sum has absolute value less than $p$, forcing it to be zero. The set $I$ is nonempty by its definition.
For sufficiency we first derive the needed <Brauer progression theorem> from the permitted <Van der Waerden theorem>. Fix <positive integers> $s,\ell$. We claim that for every number of colors $r$, a finite <integer interval> $[B(r,s,\ell)]$ forces a <monochromatic> set
$$
\{sd,\ a,a+d,\ldots,a+(\ell-1)d\},\qquad a,d>0.
$$
The case $\ell=1$ is immediate by taking $a=s$ and $d=1$. For $r=1$ and $\ell\geq2$, take $a=d=1$ in an interval of length at least $\max(s,\ell)$. Suppose the claim holds for $r-1$ and write $M=B(r-1,s,\ell)$. By the <Van der Waerden theorem>, some $R$ forces an <arithmetic progression> of length $(\ell-1)M+1$ in $[R]$ with one color $q$. Work in $[\max(R,sMR)]$ so that all needed multiples also lie in the coloring's domain. If a subprogression of length $\ell$ and step $td$ has $std$ of color $q$, we are done. Otherwise, for every $1\leq t\leq M$, the initial long <arithmetic progression> contains such a subprogression, and $std$ avoids color $q$. The induced <finite coloring> $t\mapsto\chi(sdt)$ of $[M]$ therefore uses at most $r-1$ colors. The induction hypothesis gives a <monochromatic> set $\{se,b,b+e,\ldots,b+(\ell-1)e\}$ for this induced <finite coloring>. Multiplying by $sd$ gives the desired configuration in the original coloring, with initial term $sdb$ and step $sde$. This proves the claim for $\ell\geq2$; the only use below has $\ell\geq3$.
Now suppose $\sum_{i\in I}a_i=0$ for a nonempty $I$. Choose $i_0\in I$, put $B=\sum_{i\notin I}a_i$, and put $s=|a_{i_0}|$. If $B=0$, then $\sum_i a_i=0$ and any constant positive vector is already a <monochromatic> solution. Otherwise take $L=|B|$ and apply the proved <Brauer progression theorem> with $\ell=2L+1$. All of $sd$ and $a+jd$ for $0\leq j\leq2L$ have one color. Define
$$
x_i=\begin{cases}
sd,&i\notin I,\\
a+Ld,&i\in I\setminus\{i_0\},\\
a+\bigl(L-\operatorname{sgn}(a_{i_0})B\bigr)d,&i=i_0.
\end{cases}
$$
Every coordinate is a <positive integer> in that <monochromatic> set, since $L-\operatorname{sgn}(a_{i_0})B$ is either $0$ or $2L$. Using the zero sum on $I$ gives
$$
\sum_i a_i x_i=sBd-a_{i_0}\operatorname{sgn}(a_{i_0})Bd=0.
$$
Thus both directions are proved:
$$
\boxed{(a_1,\ldots,a_n)\text{ is partition regular}\iff
\exists\,\varnothing\ne I\subseteq[n]:\ \sum_{i\in I}a_i=0.}
$$
For $n=1$ the right side is impossible, agreeing with the absence of positive solutions for a single nonzero coefficient.
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