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Use the alternating dyadic colouring excludes symmetric weighted pair sums construction:Thus consecutive dyadic intervals $[2^k,2^{k+1})$ have opposite colours. Suppose an increasing infinite sequence $X=(x_j)$ had all its distinct-index weighted sums in one colour class. Set $\delta (y)=2^{\lfloor {\log }_{2}y\rfloor +1}-y$, the distance to the next strictly larger power of $2$.

If $\delta$ is unbounded on $X$, fix a term $x$ and choose a later term $y>2x$ with $\delta (y)>2x$. Put $L=2^{\lfloor {\log }_{2}y\rfloor }$, so $y=2L-\delta (y)$. ThenThese two weighted sums lie in adjacent dyadic intervals and therefore have opposite colours.

If $\delta$ is bounded by $D$ on $X$, choose a term $x>2D$ and a later term $y>2x$. With the same $L$, we haveFor the upper bounds, $y+2x<2y<4L$ and $2y+x<\frac{5}{2}y<5L<8L$. These weighted sums also lie in adjacent dyadic intervals. In both cases they are $x_i+2x_j$ and $x_j+2x_i$ for two distinct indices, contradicting the common colour class. Hence the symmetric distinct-index conclusion is false, even for this two-colour finite colouring.