Why not a magnetic field local maximum? Reading a brief on magnetic traps I've read that you can't have a magnetic field local maximum. I believe this is linked to the divergence of B being 0 in vacuum, but I don't see why one can have a local minimum but not a maximum.
 A: For a stationary point to be a local maximum or local minimum, then the eigenvalues of the Hessian matrix must either be all negative or all positive respectively.
However, a neat trick is that the trace of the Hessian is equal to the sum of the eigenvalues and is also the Laplacian.
So, to answer the question we ask what the Laplacian of the magnetic field magnitude looks like, or more conveniently, the Laplacian of $B^2$.
$$\nabla^2 B^2 = \nabla^2(B_{x}^2 + B_{y}^2 + B_{z}^2)$$
$$\nabla^2 B^2 = \nabla \cdot \nabla(B_{x}^2 + B_{y}^2 + B_{z}^2)$$
$$\nabla^2 B^2 = 2\nabla \cdot (B_x \nabla B_x + B_y \nabla B_y + B_z \nabla B_z)$$ 
$$\nabla^2 B^2 = 2( [\nabla B_x]^2 + [\nabla B_y]^2 + [\nabla B_z]^2 + B_x \nabla^2 B_x + B_y \nabla^2 B_y + B_z \nabla^2 B_z)\ \ \ (1)$$
For a time-independent, current free situation we know that $\nabla \times {\bf B} =0$. If we take the curl of both sides
$$\nabla \times \nabla \times {\bf B} = -\nabla^2 {\bf B} + \nabla(\nabla \cdot {\bf B}) = 0$$
and therefore, because $\nabla \cdot {\bf B} = 0$, we can say $\nabla^2 {\bf B} = 0$ and we can also say that the individual components $\nabla^2 B_x = \nabla^2 B_y = \nabla^2 B_z = 0$. Using this in equation (1), we get
$$\nabla^2 B^2 = 2( [\nabla B_x]^2 + [\nabla B_y]^2 + [\nabla B_z]^2),$$
but because the square of the individual gradients must always be $\geq 0$, we can say that
$$\nabla^2 B^2 \geq 0$$.
From the discussion of the Hessian and its relationship with the eigenvalues and the Laplacian at the beginning we can thus say that whilst a local minimum in the magnitude of the magnetic field is possible because the trace of the Hessian can be $>0$, it is impossible to have a local field magnitude maximum, because it cannot be $<0$.
A: Static magnetic field is a harmonic function (its curl and divergence vanish, you can then derive that its Laplacian vanishes), so it does not have local minima or maxima.
EDIT (10/3/2015): I am grateful to the authors of the comments for their criticism. What I wrote seems to be applicable to the components of static magnetic field (their Laplacian is zero, say, if there are no charges/currents in the area). As for the magnitude of magnetic field, it looks like it can have local minimums, but not maximums (https://books.google.com/books?id=iULpBKHIbeoC&pg=PA77&lpg=PA77&dq=magnetic+field+local+maximum&source=bl&ots=N8ubx1w1QS&sig=FOyZyUjQNdPz92X3bmVlGyrDORY&hl=en&sa=X&ved=0CD8Q6AEwBGoVChMIw6Kqnc-nyAIVhIYNCh08JQAg#v=onepage&q=magnetic%20field%20local%20maximum&f=false)
A: If you place two magnets end to end with a gap in between, there will be a local minimum in the magnetic field between the poles. And inside a coil there is a local maximum.
So I don't know where you get your statement from. It seems to be wrong. Or it applies under certain conditions which you did not specify.
A: The existence of a maximum of a static magnetic field (or static electric field for that matter) in empty space can be ruled out based on the following result (see Eq. 5.63 for magnetic or Eq. 4.19 for electric fields in Jackson's Classical Electrodynamics, 3d ed.): $$\int_{r<R}\textbf{B } d^3x=\frac{4\pi R^3}{3}\textbf{B}(0)$$ or in words: the average value of a static field inside a sphere of empty space is equal to the value of the field at the center of that sphere.
Let's consider a sphere centered at the location of a suspected maximum of the field. Apparently the average value of the field must be smaller than the maximum, which contradicts the above equation and proves that a maximum cannot exist. The argument breaks down for a minimum of a field because averaging involves taking vector sum.
