Maxwell's equations, in differential form, are
$$\left\{\begin{align} \vec\nabla\cdot\vec{E}&=~\rho/\epsilon_0,\\ \vec\nabla\times\vec B~&=~\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t},\\ \vec\nabla\times\vec E~&=-\frac{\partial\vec B}{\partial t},\\ \vec\nabla\cdot\vec{B}~&=~0, \end{align}\right.$$
which are, respectively, scalar, vector, pseudovector and pseudoscalar equations. Is this purely a coincidence, or is there a deeper reason for having one of each type of equation?
If I'm not mistaken, these objects correspond to (or, at least, a correspondence can be made with) the ranks of differential forms on a 3-dimensional manifold, so I guess there might be some connection with the formulation of Maxwell's equations in terms of differential forms. If this is this case, is there an underlying physical reason that our expression of the equations turns out with one of each rank of equation, or is it a purely mathematical thing?
As a final note, I might also be totally wrong about the rank of each equation. I'm going on the contents of the right-hand side. (e.g. a "magnetic charge density" would be pseudoscalar.)
Answer
In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources.
But this is rather illusory. In relativity, the equations look quite different:
$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= 0\end{align*}$$
where $F$ is the electromagnetic field bivector. The vector derivative $\nabla$ can only increase or decrease the grade of an object by 1. Since $F$ is grade 2, the divergence equation describes its relationship with a grade 1 source term (the vector four-current $J$). The curl equation describes its relationship to a grade 3 (trivector) source term (of which there is none).
The reason the 4 Maxwell equations in 3-space come out the way we do is that we ignore the timelike basis vector, which would unify the scalar charge density with the 3-current as the four-current, as well as unify the E field with the B field as a bivector. The relativistic formulation, however, is considerably more sensible, as it correctly presents the EM field as one object of a single grade (a bivector), which can only have two sources (vector or trivector). It just so happens that the EM field has no trivector source.
What if there were trivector sources? Well, as you observe, there would be magnetic charge density (monopoles), but there would also be quite a bit more. There would have to be magnetic current as well, which would add an extra term to the $\nabla \times E$ equation to fully symmetrize things.
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