= 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 $2\times2$ matrices. Anticommutation gives $\operatorname{Tr}(AB)=0$ for distinct matrices, whereas $\operatorname{Tr}(A^2)=2$. They would therefore be four linearly independent members of the three-dimensional complex space of traceless $2\times2$ matrices, a contradiction.
$$
\boxed{N\text{ is even and }N\ge4.}
$$
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.
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