Jose has a winning strategy. Reduce the numbers modulo $2019$. If two numbers $a$ and $b$ have the same residue, Jose can delete all other numbers and calculate the sum $a-b$, which is a multiple of $2019$. Otherwise, all $2019$ numbers have different residues. Since there are only $2019$ residues, this means every residue appears exactly once. Jose can delete no numbers and calculate the sum of all numbers on the board: \[0+1+2+ \cdots +2018 \equiv \frac{2018 \cdot 2019}{2} \equiv 1009 \cdot 2019 \equiv 0\pmod {2019}\]