Alice and Bob are playing the following game:
They take turns writing on the board natural numbers not exceeding $2018$ (to write the number twice is forbidden).
Alice begins. A player wins if after his or her move there appear three numbers on the board which are in arithmetic progression.
Which player has a winning strategy?
I claim that Alice has the winning strategy.
Alice begins with the number $1014$. Let the number Bob writes be $n$. If $n\neq2018$ then Alice wins by writing $2018-n$. If $n=2018$ then Alice wins by writing $\frac{1014+2018}{2}=1516$.