36th Hellenic Mathematical Olympiad

We observe that after a move the sum of the numbers in the table have a change equal to the difference: $$(5\alpha -2\beta )+(3\alpha -4\beta )-(\alpha +\beta )=7(\alpha -\beta )$$ Therefore we conclude that after every application of a move the difference of the sum $S_{\text{new}}$ minus the sum $S_{\text{initial}}$ is a multiple of $7$, that is $$S_{\text{new}}-S_{\text{initial}}=\text{mult. }7.$$ So, the sums $S_{\text{new}}$ and $S_{\text{initial}}$ when divided by $7$ give the same remainder. Since $$S_{\text{initial}}=1+2+\cdots +2018=1009\cdot 2019\equiv 1\cdot 3(\mathrm{mod}7)\equiv 3(\mathrm{mod}7).$$ In case we find the numbers $3,6,9,\dots ,6054$, then their sum will be $$S_{\text{new}}=3+6+\cdots +6054=3\cdot S_{\text{initial}}\equiv 3\cdot 3(\mathrm{mod}7)\equiv 2(\mathrm{mod}7).$$ Therefore is not possible to find the numbers $3,6,9,\dots ,6054$ in the table, and so Mary is right.