Is the new equilibrium total pressure of this equilibrium reaction always more than its initial total pressure?
The higher pressure of $\ce{NO2}$ even after reaching equilibrium after adding some amount of $\ce{NO2}$ can be shown mathematically. 

Suppose initially before adding $\ce{NO2}$,the pressure of $\ce{NO2}$ was $p^0_{\ce{NO2}}$, and that of $\ce{N2O4}$ is $p^0_{\ce{N2O4}}$.So, the equilibrium constant is $$K_\mathrm{p} = \frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}$$ Now assume that the addition of $\ce{NO2}$ just increases the pressure instantaneously to $p_x$ but immediately after that due to dissociation to achieve equilibrium, the pressure decreases by $\ce{p_y}$ atm. 

$$\ce{2NO2 &lt;=&gt; N2O4}$$
$$p^0_{\ce{NO2}} + p_x - p_y \ce{&lt;=&gt;} p^0_{\ce{N2O4}} + \frac{p_y}{2}$$

So, the increase of pressure $\ce{N2O4}$ will be $p_y/2$ atm. So, now a new equilibrium is achieved where the equilibrium constant $K_\mathrm{p}$ will be still the same assuming that during addition of the gases, temperature is not changed( or the change is very much negligible in real cases). Thus, equating $K_\mathrm{p}$ on both the sides we have, 

$$\frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}= \frac{p^0_{\ce{N2O4}} + \frac{p_y}{2}}{(p^0_{\ce{\ce{NO2}}} + p_x - p_y)^2}$$

Taking the squares on numerator and denominator and taking square root we will find:

$$1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}}= \sqrt{1+\frac{p_y/2}{p^0_{\ce{N2O4}}}} &gt; 1$$

So, we see that $1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}} &gt; 1$. Hence it is proved that $\ce{p_x - p_y &gt; 0}$. So, the net pressure of $\ce{NO2}$ after achieving equilibrium is $p^0_\ce{NO2} + p_x - p_y &gt; p^0_{\ce{NO2}}$. So, its pressure has increased than the previous conditions even after reaching equilibrium.