Label the progressions $P_1, P_2, P_3$. If any two of 2, 3, and 4 are in the same progression, then either the difference is 1 and 1980 is obviously included or the difference is 2 and contains all naturals, in which case 1980 is again included. Therefore, we assume WLOG that 2 $\in P_1$, 3 $\in P_2$, and 4 $\in P_3$. By similar logic, we can assume that none of 4, 5, and 6 are in the same progression. This gives two cases: Case 1: 5 $\in P_2$, 6 $\in P_1$. Then 1, 3, 5, 7 $\in P_2$ and 2, 6 $\in P_1$. Therefore, 4,8 $\in P_3$ and since 4|1980, 1980 $\in P_3$. Case 2: 5 $\in P_1$, 6 $\in P_2$. Then 3, 6 $\in P_2$. And since 3|1980, it follows that 1980 $\in P_2$.