Until the first meeting point, the first cyclist travelled the distance $S_1 = v_1$, while the second cyclist travelled $S_2 = v_2$. Therefore, the distance between towns $A$ and $B$ equals $v_1 + v_2$. There are two possible cases.

Case 1. Before their second meeting, both cyclists reached their destination town, and turned around. Suppose their second meeting occurred at the distance $S_3$ from town $B$ and time $t_2$ later after the first meeting (fig. 22). Then, we get the equations

Fig. 23
$$v_2 + S_3 = v_1 t_2 \text{ and } v_1 + (v_1 + v_2 - S_3) = v_2 t_2.$$
By adding these equations, we obtain
$$v_2 + S_3 + 2v_1 + v_2 - S_3 = (v_2 + v_1)t_2 \Rightarrow t_2 = 2.$$

Case 2. Before their second meeting, the first cyclist reached destination town $B$, turned around and reached the second cyclist before he/she reached town $A$ (fig. 23). Suppose their second meeting occurred at the distance $S_3$ from town $B$ and time $t_2$ later after the first meeting (fig. 4). Then, we get the equations
$$S_3 - v_2 = v_2 t_2 \text{ and } v_2 + S_3 = v_1 t_2.$$
Taking the difference of these equations, we obtain
$$v_2 + S_3 + v_2 - S_3 = (v_1 - v_2)t_2 \Rightarrow t_2 = \frac{2v_2}{v_1 - v_2}.$$

Now, the only thing left is to figure out which of the two cases occurs for which $v_1, v_2$. Case 1 happens when the first cyclist reaches $A$ later than the second cyclist, i.e. $\frac{v_2+v_2+v_1}{v_1} > \frac{v_1}{v_2} \Rightarrow 2v_2^2 + v_1v_2 > v_1^2$. Let us denote $x = \frac{v_1}{v_2}$, then it must satisfy the equation
$$x^2 - x - 2 < 0 \Rightarrow (x+1)(x-2) < 0 \Rightarrow x < 2, \text{ i.e. } v_1 < 2v_2.$$