Let's say we have two modes, with the following labeling of occupation number states:
$ \lvert \Psi \rangle = \begin{pmatrix} 0,0 \\ 0,1 \\ 1,0 \\ 1,1 \end{pmatrix} $
An example of (what I assume to be) fermionic creation operators for the two modes is
$\hat a_1^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \quad \hat a_2^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$
These operators obey full anti-commutation relations.
$\{\hat a_1,\hat a_1^\dagger\} = \{\hat a_2,\hat a_2^\dagger\} = 1$
$a^\dagger_1 a^\dagger_1 = a^\dagger_2 a^\dagger_2 = 0$
$\{\hat a_1,\hat a_2^\dagger\} = \{\hat a_1,\hat a_2\} = 0$
If we don't include the ($-$) sign, then operators corresponding to the same mode still anti-commute, but those corresponding to different modes commute.
$\hat b_1^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \quad \hat b_2^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$
$\{\hat b_1,\hat b_1^\dagger\} = \{\hat b_2,\hat b_2^\dagger\} = 1$
$b^\dagger_1 b^\dagger_1 = b^\dagger_2 b^\dagger_2 = 0$
$[\hat b_1^\dagger,\hat b_2^\dagger] = [\hat b_1^\dagger,\hat b_2] = 0$
It looks like we started constructing a boson Fock space, but only included states for which the occupation numbers are 0 or 1. Is there some reason these operators aren't suitable, other than the observation that all elementary particles are either fermions or bosons? Are there any quasi-particles in condensed matter physics that behave like this?