Hard particles and soft particles
It is a sloppy usage, hard particles are the ones coming from "hard scattering", i.e. deep inelastic scattering, and are particles with a lot of energy. Soft particles are the ones with low energy and are the spectators or debris of the strong interaction.
The distinction is meaningful in proton proton or proton antiproton scattering, see this link for example.
It's a shorthand for describing the particle's kinetic energy, by anticipating the sorts of collisions that a particle will produce. More energetic particles will "hit harder" and less energetic particles will "hit softer".
annav's answer gives some examples in from high-energy physics, but low-energy people use this terminology as well. Thermal neutrons that capture on cadmium produce "hard" 8--10 MeV gamma rays, while neutrons on boron produce "soft" half-MeV gamma rays. The one-syllable description of the energy of the photon is an efficient way to remind yourself that you have different shielding requirements for different targets. (Don't laugh at me, high-energy people, for thinking of a 10-MeV photon as "hard"; since that's typical nucleon separation energy that's basically as much as you can get from an intact nucleus at rest.)
I had the experience as a grad student of putting some absorber in a neutron beam that was better at eating slower neutrons than faster ones. I showed my mentor a before-and-after plot of the neutron spectrum, where there were fewer neutrons at every energy after the absorber than before, and he noticed the change in shape and said "Oh, okay, the absorber has hardened the spectrum." That's a pretty typical usage in my experience.
The distribution of particle energies is expressed in terms of a spectrum $dN/dE$ where $N$ is the number of particles with energy $E$. When comparing two populations of particles, the one with a greater proportion of high energies is termed as "harder". So for example if the spectra are well parameterised by a power law ($dN_i/dE\propto E^{-\alpha_i}$) then population 1 will have a harder spectrum than population 2 if $\alpha_1<\alpha_2$.