There is a Minimum

$ad-bc$ is an integer due to closure property of integers over subtraction and multiplication. Since $a,b,c,d$ are positive, by cross multiplying we get $ad-bc>0$ . So its least value can be $1$ . For example $\frac {1}{2}>\frac {1}{3}$ . Here we can clearly say, that $ad-bc$ attains value $1$ . Hence proved.

$\dfrac{a}{b} > \dfrac{c}{d} \\ ac > db \\ ac - bd > 0 \\ \implies \mathrm{min}(ab-bd) = 1$