1. If the system of inequalities $\left\{\begin{array}{l}x-1000 \geqslant 1018, \\ x+1 \leqslant a\end{array}\right.$ has only a finite number of real solutions, then the value of $a$ is $\qquad$.
二、1.2 019.
The solution set of the system of inequalities is
$$
2018 \leqslant x \leqslant a-1 \text {. }
$$

If $a-1>2018$, then there are infinitely many real numbers between $2018 \sim a-1$, which does not meet the requirement.
Therefore, $a-1=2018 \Rightarrow a=2019$.