Minimum sum

By Ptolemy's Inequality, $AB \times CD+AD \times BC \geq AC \times BD$

Let $AC$ and $BD$ meet at $P$ . Let $PA, PB, PC, PD$ be $a,b,c,d$ . Let $\theta=\angle APB$ be the angle between $AC$ and $BD$ .

$2017=\frac12 (ab+cd)\sin{\theta} + \frac12 (ac+bd)\sin{(180^\circ - \theta)}$ $=\frac12(ab+bc+cd+da)\sin{\theta}$

Hence $AC \times BD=(a+c)(b+d)$ $=ab+bc+cd+da$ $=\dfrac{2*2017}{\sin{\theta}}$ $\geq \boxed{4034}$

Equality holds when the quadrilateral is cyclic and has perpendicular diagonals.

For a quadrilateral, a square has greatest area and least perimeter . And AB * CD + AD * BC = AC * BD=AC * AC.
Sides of a square area 2017 = $\sqrt{2017}$ . Diagonal AC=BD= $\sqrt{2*2017}$ .
So AB * CD + AD * BC = AC * BD=AC * AC = 2 * 2017 = $\Large \color{#D61F06}{4034}.$

The only way to minimize the side lengths is if the quadrilateral is both equilateral and equiangular. This makes a square. AB * CD = AD * BC = 2017 *2 = $\boxed{4034}$ .