Put $A$ at the origin and $E = (20, 0);$ the right angles at $A$ and $E$ make $AB$ and $ED$ vertical.

The $45^\circ$ angles at $B$ and $D$ force $C = (14, AB - 14)$ and $DE = AB - 8;$ the shoelace formula turns the area condition into a congruence for $AB.$

Place $A = (0, 0)$ and $E = (20, 0)$ with the pentagon above line $AE,$ and write $h = AB.$ The right angles at $A$ and $E$ make $AB$ and $ED$ vertical: $B = (0, h)$ and $D = (20, k)$ with $k = DE.$ At $B$ the side $BC = 14\sqrt{2}$ makes a $45^\circ$ angle with the downward ray $BA,$ heading into the pentagon, so $C = (14, h - 14).$ Similarly at $D,$ the side $DC$ makes a $45^\circ$ angle with the downward ray $DE,$ so $C = (20 - s, k - s)$ where $s = \frac{DC}{\sqrt{2}}.$ Matching coordinates gives $s = 6$ and $k = h - 8.$ The interior angle at $C$ is then the reflex angle $270^\circ$ (angle sum $90 + 45 + 270 + 45 + 90 = 540$), and $DE = h - 8 \lt AB$ automatically.

The shoelace formula on $A(0,0),$ $B(0,h),$ $C(14, h-14),$ $D(20, h-8),$ $E(20, 0)$ gives area $$\begin{aligned} [ABCDE] &= \frac{1}{2}\big|{-14h} + (-6h + 168) \\ &\qquad {}+ (-20h + 160)\big| \\ &= 20h - 164. \end{aligned}$$ The condition that $20h - 164$ be divisible by $16$ reduces to $4h \equiv 4 \pmod{16},$ that is, $h \equiv 1 \pmod 4.$ For $C$ to lie strictly on the same side of line $AE$ as $B$ and $D,$ we need $h \gt 14.$

So $h$ runs over $17, 21, 25, \ldots, 2025,$ which is $\frac{2025 - 17}{4} + 1 = 503$ values.