Summary
At sufficiently low volumes, the excluded volume effect causes the van der Waals (VDW) pressure of all gases to exceed the ideal gas (IG) pressure. At higher volumes, the VDW pressure may or may not cross below the IG pressure. This depends on the temperature, and the relative size of the VDW a and b terms. If the temperature is sufficiently low, and if the a term is sufficiently large relative to the b term, you will see a crossover. Otherwise, you won't.
Notes:
(1) The OP asks for a discussion of VDW gases, and thus most of this analysis concerns the predictions of the VDW model. However, the curves presented in the OP's screenshot are for real gases and thus, at the end, I discuss their behavior.
(2) Below the VDW critical temperature (which equals $\frac{8a}{27bR}$, and is thus gas-specific), the VDW isotherms (the curves on a $p-V$ graph) begin to show oscillations. Those oscillations are beyond the scope of this question, but they should be noted for completeness. This answer assumes we are above the VDW critical temperature for each gas.
The VDW model
The VDW model assumes the gas molecules are hard spheres that interact only attractively. Unlike the case with the attractive forces, there are no long-range repulsive forces in the VDW model. Indeed, there are no repulsive forces at all between the gas molecules (except insofar as they are hard spheres that can't interpenetrate).
Instead, the molecules are assumed to have a physical volume, and the effect of this volume is to reduce the free volume left for the gas. The volume of one mole of molecules (or atoms) is given by the VDW "b" term.
This should help illustrate how the physical volume of the gas particles affects the pressure:
Suppose you are at a volume where the hard spheres themselves take up half the container. If there were no attractive forces, the pressure of the VDW gas would be double that of the ideal gas, because only half as much free volume is left for the VDW gas, such that p will be a monotonically decreasing function of V.
Physical explanation of difference between pressure of VDW gas and ideal gas
At very low volume (low relative to the volume taken up by the hard spheres), the molecular volume term dominates, and thus p_VDW > p_IG (where IG = ideal gas).
What happens at higher volumes, however, is gas- and temperature-specific. And thus whether or not there is a crossover is gas- and temperature-specific.
Case I: Lower temperature
At a sufficiently high volume, the pressure-increasing effect of molecular volume becomes less than the pressure-reducing effect of the attractive force, and thus p_VDW < p_IG. Hence there is a point at which the VDW and IG curves cross over.
Again, let me emphasize that what constitutes "lower" vs. "higher" temperature is gas-specific. For instance, while $T = 273 \, \ce{K}$ is in the low-temperature regime for $\ce {N2(g)}$, it's in the high-temperature regime for Ne. Thus, at this temp, you would see a crossover for $\ce {N2(g)}$, but not for Ne.
More specifically, $\ce {Ne(g)}$, which has a small a value relative to b, loses its crossover behavior above $149.8 \, \ce{K}$, while $\ce {N2(g)}$ doesn't lose its crossover behavior until the temperature exceeds $425.8 \, \ce{K}$.
Case II: Higher temperature
As explained above, as the volume increases, the effect of molecular volume decreases. However, the intermolecular spacing also increases, lessening the average attractive force between the particles. Further at, high temperatures the kinetic energy of the gas is higher, thus making the effect on pressure of intermolecular attraction relatively less. If the attractive force is sufficiently low, and the temperature is sufficiently high, then the attractive force is never able to overcome the molecular volume effect. Hence, for these gases at these temperatures, p_VDW > p_IG at all volumes.
Note: the above discussion is referring to the VDW model, not necessarily to how real gases actually behave.
Mathematical explanation of difference between pressure of VDW gas and ideal gas
We can also see this from the mathematical structure of the VDW equation:
$$p=\frac{RT}{V_m-b}-\frac{a}{V_m^2}$$
In the limit as $V_m \rightarrow b$, $(V_m-b)\rightarrow 0$, causing the first term to dominate over the second.
As $V_m$ gets large, whether the attraction can win out over the excluded volume is determined by the magnitude of T, and the relative values of a and b.
In the limit as $V_m \rightarrow \infty$, for all gases, the VDW equation reduce to the ideal gas equation:
$$p=\frac{RT}{V_m}$$
Plots, Case I, lower T ($\ce{N2(g)}$ at $\pu{273K}$)
Comparative pressure vs. volume plots of VDW and ideal gas
Now let's take a look at what the actual $p-V$ graphs look like, using $\ce {N2(g)}$ as an example. The VDW a and b terms for $\ce {N2(g)}$ are:
$$a=1.37 \ce{ \frac{bar L^2}{mol^2}}$$
$$b=0.0387 \ce{ \frac{L}{mol}}$$
Assuming $T=273 \,\ce{K}$, we can determine crossover point by setting pIdeal = pN2VDW (the VDW pressure for $\ce{N2(g)}$), and solving for $p$. The crossover occurs at $V = 0.108 \,\ce{ L}$. But if we plot pIdeal and pN2VDW from about 0.1 L to 10.0 L, we find they are nearly on top of each other:

Thus to see the crossover, we need to focus on the region around $V = 0.108 \,\ce{ L}$:

Plot of p_VDW–p_ideal vs volume, using $\ce{N2 (g)}$ as the VDW gas
We can further improve our understanding by plotting the difference between pIdeal and pN2VDW vs $V$. This allows us to see, directly, how the deviation between the VDW equation and the ideal gas equation changes with volume.
Let's start at high volume, and move to the left. At high volume, the attractive term dominates. As the volume decreases, the molecules get closer together, so the attractive term has even more effect, increasing the difference between pIdeal and pVDWN2. This continues until $V=0.154 \,\ce{L}$ [I determined this minimum by setting the derivative of (pN2VDW - pIdeal) equal to zero, and solving for $V$.] Beyond this, the increase in pressure due to the diminishing free volume becomes greater than the decrease in pressure due to the increasingly strong attractive forces. Eventually, at $V=0.108 \,\ce{L}$ the effect of the hard sphere volume just balances that of the attractive forces, and the curve crosses above the zero line.

Plots, Case II, higher T ($\ce{N2(g)}$ at $\pu{500 K}$)
Assuming $T = 500 \,\ce{K}$, if we solve for the crossover point as we did above, we get a non-physical answer of $V = -0.222 \,\ce{ L}$. This indicates the curves don't cross over. We can see this from the plots:


Real Gases
The easiest way to see how the isotherms for real gases compare with those for an ideal gas is to look at the measured compressibility factor, $Z$:
$$Z=\frac{p V_b}{R T}$$
$Z = 1$ for an ideal gas. Thus, for a given $V_b$ and $T$, $Z>1 \implies p_{real} >p_{ideal}$, and visa-versa.
From https://en.wikipedia.org/wiki/Compressibility_factor#/media/File:Z_Overview.png, we have the following diagram for $\ce{N2(g)}$. It appears to show behavior similar to what the VDW equation predicts— at lower temperatures, we see a crossover from
$p_{real} > p_{ideal}$ at high $p$ (low $V$), to $p_{real} < p_{ideal}$ at low $p$ (high $V$).
However, as the temperature increases, $p_{real}$ reduces its tendency to dip below $p_{ideal}$ (see red curve). However, a direct examination of the numerical data would be needed to determine if, at high $T$, there is no crossover at all.
