60th Belarusian Mathematical Olympiad

It is easy to see that if either $1$ or $5$ numbers remain on the blackboard, then the player who must move wins, but if $2$ or $4$ numbers remain, then this player loses. So we will solve the problem moving backward. We write all numbers from $1$ to $80$ and mark them with "+" or "-". If $k$ numbers remain on the blackboard before the move of a player and he can win, then we write $+k$, otherwise we write $-k$. We have the following table:

+1	-2	+3	-4	+5	-6	+7	+8	+9	+10	+11	+12	-13
1	-	1	-	5	-	1	8	5	8	5	8	-

+14	-15	+16	-17	+18	-19	+20	+21	+22	+23	+24	+25	-26	...
1	-	1	-	5	-	1	8	5	8	5	8	-	...

The second row of the table contains the possible winning moves (to win the player can erase so many numbers that after his move the quantity of the numbers remained on the blackboard would be marked with "+"). We see that the signs in the first row of the table are repeated with the period equaled $13$ (it is easy to prove by induction).

Since $80$ leaves remainder $2$ when divided by $13$, it follows that $80$ and $2$ are marked with the same sign, i.e. "-". Thus the number $80$ is the losing number of the numbers for the beginning player. To win Jerry can erase so many numbers as it is written in the second row of the table.