Can a particle have a decay width approaching its mass? Very short-lived particles have half-lives measured in terms of decay width, with the half-life calculated from the energy-time uncertainly principle.
If a particle has a decay width that approaches the mass of the particle, it never really gets a chance to "exist". No one will ever see it.
But could such a particle (or quasi-particle in condensed matter) "exist" in the sense that it has shows up as a virtual particle in Feynman diagram calculations, with a mass, spin, charge, etc? 
The best candidate of such a particle I can think of is the Planck Particle, but quantum gravity isn't renormalizable.
 A: Well, "no one will ever see it" is excessively harsh. Very broad resonant states are surveyed with partial wave phase-shift analysis, although this sort of thing is more in vogue in nuclear physics these days than in particle physics. However, the fabled scalar σ resonance underlying the eponymous model of chiral symmetry breaking in effective lowest-energy QCD has made its way in and out of the PDG, and, today, it is in . 
It has the quantum numbers of the vacuum: it is spinless, neutral, $I^G(J^{PC})=0^+(0^{++})$ : 
the celebrated $f^0(500)$ with $\Gamma\approx 400-700$MeV, give or take...
Its Breit-Wigner distribution is now
$$
 f(E) = \frac{cM^3}{\left(E^2-M^2\right)^2+M^4}~, 
$$
where the normalization constant c is 4π/3, I think, from 
$
\int_0^\infty  \!\!dE ~~f(E) =1.
$ 
In these conventions, $f(0)=f(\sqrt{2} M) = f(M)/2$.
In any case, even though the relativistic Breit-Wigner arises from the propagator of the unstable particle, you are not always invited to do perturbation theory with it. For the σ this corresponds to an effective theory of the ferociously non-perturbative chiral symmetry breaking process, normally not even renormalizable. So the urge to dive into the Planck mass for illustration might be  misplaced.  
