Mathematica competitions in Croatia

In each step the sum of all numbers on the blackboard changes by $$((3a-b)+(13a-3b))-(a+b)=15a-5b=5(3a-b).$$ This difference is divisible by $5$, i.e. the sum of all numbers written on the board in each step gives the same remainder when divided by $5$. At the beginning, the sum of all numbers on the board is $$1+2+3+\cdots +2012=\frac{2012\cdot 2013}{2}=1006\cdot 2013$$ and we have $$1006\cdot 2013\equiv 1\cdot 3\equiv 3(\mathrm{mod}5).$$ On the other hand, $$2+4+6+\cdots +4024=2\cdot (1+2+3+\cdots +2012)\equiv 2\cdot 3\equiv 1(\mathrm{mod}5).$$ In conclusion, it is impossible to get the numbers $2$, $4$, $6$, $\dots$, $4024$.