= Solution
Let $\rho$ be the <cyclic logarithmic coloring> from part (iii). Define a <finite coloring> of the <positive integers> by
$$
\chi(n)=\rho(\log_2 n)\quad(n\geq2),\qquad \chi(1)=0.
$$
The inner <logarithm> is at least $1$ whenever $n\geq2$, so this is defined everywhere. Suppose an increasing infinite <sequence> $(x_i)$ made every product with distinct indices <monochromatic>. Fix an index $i$ with $x_i\geq2$ and write $a=\log_2 x_i>0$. The <sequence> is unbounded, so choose $j>i$ with $b=\log_2 x_j\geq28a$. Both ordered products are among the supposedly <monochromatic> values, but their <logarithms> satisfy
$$
\frac{\log_2(x_i x_j^2)}{\log_2(x_j x_i^2)}
=\frac{a+2b}{2a+b},\qquad
\boxed{1.9\leq\frac{a+2b}{2a+b}<2.}
$$
Indeed, the first inequality is equivalent to $b\geq28a$, and the second uses $a>0$. Part (iii) therefore gives $\chi(x_i x_j^2)\ne\chi(x_j x_i^2)$, a contradiction. Thus \b[the assertion with all distinct ordered indices is false]. The obstruction needs an infinite unbounded <sequence>; it does not assert that every pair of different <positive integers> gives different colors.
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