If I have the energy of a free electron gas in three dimensions centered at energy $E_0$, $$E_n(k)=E_0+\frac{\hbar^2}{2}\left(\frac{k_x^2}{m_x}+\frac{k_y^2}{m_y}+\frac{k_z^2}{m_z}\right),$$ I can prove that the density of states in the Fermi level is $$d_n(E_F)=\frac{3}{2}\frac{n}{E_F-E_0},$$ where $n$ is the number of electrons per unit volume. Now, if all the $m_i$ are negative, I would have $E_F<E_0$ and a negative density of states? Is that right? I've read that is called a van Hove singularity of the fourth type. Can anyone help me with a physical interpretation of the phenomenon?
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I think you have a sign error; your numerator should be $(E_{0}-E_{F})$. Be careful in the derivation to keep track of the sign of $m$ and treating the $\delta$-function function composition rules carefully.
Then, for $E_{F} < E_{0}$, the density of states is finite and positive. For $E_{F} = E_{0}$, the DOS is infinite because of the van Hove singularity. For $E_{F} > E_{0}$ the DOS is actually zero at the Fermi energy, because the Fermi energy lies above the band (which is sloping downwards because of the negative $m$'s), and the formula no longer applies.
