Does a bivariate c.d.f. have this property? Let $F(\cdot,\cdot)$ denote the c.d.f. of a bivariate random vector with the support $\mathbb{R}^2$. 
Suppose that for some constants $\delta_1>0, \Delta_1>0$ and $\delta_2>0, \Delta_2>0$ such that $\delta_1\geq\Delta_1$ and $\delta_2\leq\Delta_2$ it holds that 
$$F(z_1+\Delta_1,z_2+\Delta_2) - F(z_1,z_2+\Delta_2) -F(z_1+\Delta_1,z_2)+F(z_1,z_2) = \\  F(z_1+\delta_1,z_2+\delta_2) - F(z_1,z_2+\delta_2) -F(z_1+\delta_1,z_2)+F(z_1,z_2) $$
for all $(z_1,z_2) \in \mathbb{R}^2$. 
Does it necessarily imply that $\delta_1=\Delta_1$ and $\delta_2=\Delta_2$ regardless of the specific form of the c.d.f. $F(\cdot,\cdot)$? I am happy to impose additional properties on $F$ if this would help. 
I think I can establish this property when the support of the random vector is bounded but ultimately I am interested in the general case. 
 A: This visual argument isn't a complete proof but perhaps a step in the right direction.
Visually, the condition
$$
F(z_1+\Delta_1,z_2+\Delta_2)-F(z_1,z_2+\Delta_2)-F(z_1+\Delta_1,z_2)+F(z_1,z_2)=\\F(z_1+\delta_1,z_2+\delta_2)-F(z_1,z_2+\delta_2)-F(z_1+\delta_1,\delta_2)+F(z_1,z_2)
$$
corresponds to the diagram below where the blue rectangle and the orange rectangle have to contain equal probability mass.

Since the two rectangles always overlap in this way when $\delta_1 \ge \Delta_1$ and $\delta_2 \le \Delta_2$, we know that the two rectangles below (modified from the two above) have to contain equal probability mass since we just removed the shared rectangle.

We now show that
\begin{equation}
(\delta_1 - 2\Delta_1)(\Delta_2-2\delta_2) > 0 \qquad... (1)
\end{equation}
We demonstrate the basic proof by contradiction using $\delta_1 = 2\Delta_1$ and $\Delta_2 = 2\delta_2$, which does not satisfy (1) above.

Note that the two rectangles have the same area as well as containing the same probability mass. If we shift this diagram up and left and superimpose it on itself, we get the figure below. This operation corresponds to considering the original constraint for $z_1' = z_1-\Delta_1$ and $z_2' = z_2+\delta_2$.

This figure shows that the probability mass contained in region $A$ equals that contained in regions $B$ and $C$. However since each of these regions contains positive probability mass (since the distribution has support on $\mathbb{R}^2$), and since we can repeat this replication process indefinitely, this means the total probability mass of the sum of all these regions would scale without bound. This is impossible since the total probability mass of a distribution is 1. We conclude that ($\delta_1 = 2\Delta_1$ and $\Delta_2 = 2\delta_2$) is impossible.
A similar argument holds for each of the figures below. In each case either $P(A) > P(B) > P(C)$ or $P(A) < P(B) < P(C)$, which means the amount of probability mass in these equal-area regions grows without bound as we replicate the figure either up-and-left or down-and-right.


This argument does not work when (1) holds since we get the following replication diagram where the blue and orange rectangles are not subsets of each other in either direction.

I'm not yet sure how to proceed from here.
