 Solution

The zero trace and unit eigenvalues imply equal numbers of $+1$ and $-1$ eigenvalues, so $N$ is even. A two-dimensional representation is impossible: all four matrices ${\alpha }_1,{\alpha }_2,{\alpha }_3,\beta$ would be traceless $2\times 2$ matrices. Anticommutation gives $\mathrm{Tr}(AB)=0$ for distinct matrices, whereas $\mathrm{Tr}(A^2)=2$. They would therefore be four linearly independent members of the three-dimensional complex space of traceless $2\times 2$ matrices, a contradiction.The usual irreducible $3+1$ dimensional Dirac representation has $N=4$. Larger representations can contain multiple copies; the requested lower bound is not itself a proof of irreducibility.