It is easy to write down all the pairs of the numbers, that will successively appear on the board:
$$a, b \rightarrow b+a, b-a \rightarrow 2a, 2b \rightarrow 2(b+a), 2(b-a) \rightarrow 2a^2, 2b^2 \dots$$

As we see, after the appearance of the first four numbers: $a, b, b+a, b-a$ – all the other numbers that may appear on the board are even. Since 2019 is an odd number, then the number 2019 must appear among these first four numbers.

Case 1: $b = 2019$.

Case 2: $a = 2019 < b$.

Case 3: $b-a = 2019 < b$.

Case 4: $b+a = 2019$, then $2b > b+a = 2019 \Rightarrow 2b \ge 2020 \Rightarrow b \ge 1010$. Thus the smallest possible value of $b$ is $b=1010$. Then if $b=1010$ and $a=1009$ the number 2019 can appear on the board.