Is resistance proportional to resistivity: $R=\textrm{const}\cdot\rho\;?$ Suppose we have a resistor in a strange shape, filled with a medium of resistivity $\rho$, assuming only Maxwell's equations apply, is it true that resistance $R$ is proportional to $\rho$, even for very low resistivities?
 A: If your question is: does changing the resistivity ($\rho$), while keeping the same the shape of the resistor, linearly affect the total resistance? Then the answer is yes. The "const" in your formula is then determined by the shape of the resistor.
If your irregularly shaped resistor is much longer than it is wide, and the width does not change fast, its resistance is approximated by:
$$ R = \int_0^l \frac{\rho}{A(l')}~\mathrm dl'$$
Where $\rho$ is resistivity, $A$ is the cross-sectional area and $l$ the length of the resistor. $A$ is a function of $l'$ and can't be taken out of the integral. Because $\rho$ is constant (and only if it is, this does not apply if your resistor is made of different materials), it can be taken out of the integral, so $R$ scales linearly with $\rho$.
Edit:
The original equations published by Maxwell included Ohm's law, which is in vector form: $\vec E = \rho \vec J$ (for consistency I used $\rho$ as resistivity, it should not be confused with charge density).
Then, because
\begin{align}I &= \iint_{A} \vec J \cdot ~\mathrm d\vec a \approx JA\\
\Delta U &= \int_{\vec x_1}^{\vec x_2} \vec E \cdot ~\mathrm d\vec x = \int_{\vec x_1}^{\vec x_2}\rho \vec J \cdot ~\mathrm d \vec x \approx \int_{x_1}^{x_2}\frac{\rho I}{A} ~\mathrm dx\\
R &= U/I \approx \int_{x_1}^{x_2} \frac{\rho}{A}~\mathrm dx\end{align}
We can divide by I because I must be the same everywhere in the resistor and does not depend on $x$! Because $\rho$ does also not depend on $x$, it can be taken outside the integral.
