1. Given that $a$ and $b$ are integers, $\frac{127}{a}-\frac{16}{b}=1$. Then the maximum value of $b$ is $\qquad$ .
$$
\text { Two, 1.2 } 016 .
$$

The original equation is transformed into $b=\frac{16 a}{127-a}=\frac{127 \times 16}{127-a}-16$. When and only when $127-a=1$, $b$ reaches its maximum value, at this point, $b=127 \times 16-16=2016$.