7. Given two unequal positive integers $a$ and $b$ that satisfy $a^{2}-b^{2}=2018-2 a$.
Find the value of $a+b$.
7. From the given, we know
$$
\begin{array}{l}
2019=a^{2}+2 a+1-b^{2} \\
=(a+1+b)(a+1-b) .
\end{array}
$$

Notice that, $2019=3 \times 673$,
$$
a+1+b>a+1-b>1 \text {. }
$$

Therefore, $a+1+b=673 \Rightarrow a+b=672$.