Solution:

All four bodies lie on one line exactly when every pair of planets is collinear with the star, i.e. when each pair's angular positions differ by a multiple of $180^\circ$ — half a revolution. In $n$ years the planets complete $\frac{n}{60},$ $\frac{n}{84},$ and $\frac{n}{140}$ revolutions, so the pairwise differences are $$\frac{n}{60} - \frac{n}{84} = \frac{n}{210}, \qquad \frac{n}{84} - \frac{n}{140} = \frac{n}{210}, \qquad \frac{n}{60} - \frac{n}{140} = \frac{n}{105}.$$

We need $\frac{n}{210}$ and $\frac{n}{105}$ to be multiples of $\frac{1}{2}.$ The first requires $n$ to be a multiple of $105,$ and any such $n$ makes $\frac{n}{105}$ an integer. The smallest positive choice is $n = 105.$