The curvature of the Earth is not enough to really 'curve' any object that's city-sized or smaller. The reason for this is that the radius of the Earth is really big.[citation needed]
To make this more precise, say you make your chair one metre wide, and you have the edges pointing towards the centre of the Earth. You can then form an isosceles triangle between the chair's top and the Earth's centre, whose sides are $w=1\,\text m$ and $R_\oplus=6,370\,\text{km}$, so that the angular deviation from parallel edges is, to very good accuracy,
$$
\delta\theta=2\arctan\left(\frac{w}{2R_\oplus}\right)\approx\frac{w}{R_\oplus}\approx2.7\times10^{-9}{}^\circ.
$$
If you want to get the curvature of the Earth to be noticeable, you need objects and distances that are comparable with $R_\oplus$. As you note, inter-city distances are often on this scale.
Thus, one thing that I would suggest is having a 'roadsign' that points to where the other cities actually are. For example, the distance from London to Sydney is, as the crow flies, $d=3,465\:\mathrm{km}$, and this means that they subtend an angle of
$$
\theta=\frac d{R_\oplus}\approx 0.54\:\mathrm{rad}=31^\circ.
$$
This means that if you're in London and want to point straight to New York, you need to point west-by-northwest but also about thirty degrees below the horizon.

That is, I think, your best shot.