Solution:

If a square of side $t$ has its vertices on the sides of a concentric square of side $s$ and is tilted by angle $\theta,$ each side of the outer square is split into pieces $t\cos\theta$ and $t\sin\theta,$ so $s = t(\cos\theta + \sin\theta).$ This applies to $IJKL$ (side $t$) in $ABCD$ (side $s$) with some angle $\theta.$ Since $\overline{EF} \parallel \overline{AB},$ square $EFGH$ (side $u$) makes the same angle $\theta$ with $IJKL,$ so also $t = u(\cos\theta + \sin\theta).$

Hence $\frac{s}{t} = \frac{t}{u},$ so the three areas form a geometric progression: the area of $EFGH$ equals $\frac{T^2}{2016},$ where $T$ is the area of $IJKL.$ As $\theta$ ranges over $\left(0^\circ, 90^\circ\right),$ the factor $(\cos\theta + \sin\theta)^2$ takes every value in $(1, 2],$ so $T = \frac{2016}{(\cos\theta + \sin\theta)^2}$ takes every value in $[1008, 2016)$ ($\theta = 0^\circ$ is excluded because $EFGH$ is smaller than $ABCD$). For $\frac{T^2}{2016}$ to be an integer, $2016 = 2^5 \cdot 3^2 \cdot 7$ must divide $T^2,$ which forces $2^3 \cdot 3 \cdot 7 = 168$ to divide $T.$

The multiples of $168$ in $[1008, 2016)$ run from $1008$ to $1848,$ and each is attained by an appropriate $\theta,$ with the area of $EFGH$ then a positive integer less than $2016.$ The difference is $1848 - 1008 = 840.$