= Solution
Let $c_1,\ldots,c_n$ be the columns of the rational <partition regular matrix> $A$, and clear denominators so that they are integer vectors. We use the <P-adic columns lemma>. For a large <prime number> $p$, color each positive integer by a sufficiently long initial block of the unit part of its <P-adic valuation>, together with its valuation modulo the block length. Partition regularity supplies a monochromatic $x=(x_1,\ldots,x_n)$ with
$$
\sum_{i=1}^n x_i c_i=0.
$$
Group the indices according to the successive $p$-adic orders of the $x_i$. At the lowest order, division by the common power of $p$ and reduction modulo the chosen large power shows
$$
\sum_{i\in B_1}c_i=0.
$$
Comparing the next nonzero blocks of base-$p$ digits shows successively that
$$
\sum_{i\in B_j}c_i\in
\operatorname{span}_{\mathbb Q}\{c_i:i\in B_1\cup\cdots\cup B_{j-1}\}
\qquad(j>1).
$$
For completeness, these congruences may be made exact by taking the digit block longer than every determinant and coordinate formed from the fixed columns: a nonzero such integer cannot be divisible by the resulting power of $p$. There are only finitely many ordered partitions of $[n]$, so passing through arbitrarily long blocks leaves one partition satisfying all the displayed identities. This is precisely the <columns property>.
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