= Solution
Use $\mathbb N=\{1,2,\ldots\}$. First recall the <Ramsey theorem for r-sets> in its infinite form: every finite colouring of the $r$-element subsets of an infinite countable set has an infinite homogeneous subset. Here is a proof, so the combinatorial input is explicit.
For $r=1$, this is the <infinite pigeonhole principle>. Suppose it holds for $r$. Given a colouring of $(r+1)$-sets, choose a first point $a_1$. Colour the $r$-sets in the remaining tail by adjoining $a_1$, and use the induction hypothesis to obtain an infinite tail on which that colouring is constant, with colour $d_1$. Choose $a_2$ from that tail and repeat, always thinning the unused tail. This produces increasing $a_i$, nested infinite reservoirs containing all later selected points, and colours $d_i$ such that every $(r+1)$-set of selected points whose least point is $a_i$ has colour $d_i$. Infinitely many $d_i$ are equal. Keeping the corresponding points gives an infinite homogeneous subset. This is a <successive thinning proof of the infinite Ramsey theorem>.
Now let $\chi$ be the given finite colouring of positive integers. Colour each unordered pair $\{a,b\}$, with $a<b$, by $\chi(a+2b)$. Apply the proved theorem with $r=2$, and enumerate its homogeneous set increasingly as $x_1<x_2<\cdots$. Then
$$
\boxed{\chi(x_i+2x_j)=\gamma\quad\text{for all }i<j}
$$
for one fixed colour $\gamma$. This proves the requested monochromatic pattern without requiring the $x_i$ themselves to have that colour.
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