It is a standard fact which is explicitely or implicitely in every reference on supersymmetry. See for example the beginning of "Supersymmetry and supergravity" by Wess and Bagger.

The question is implicitely about spacetime dimension equal to 4. In this case, the smallest spinor representation is a Weyl spinor, with two complex components. To have $N$ supersymmetries means that we consider $N$ Weyl spinors $Q_{\alpha}^i$, $i=1,...,N$ ($\alpha=1,2$ is the Weyl spinor index), as fermionic generators of the supersymmetry algebra. The supersymmetry algebra is a super Lie algebra, extension of the Poincaré Lie algebra. Without central charges, it contains the relations

$\{ Q_\alpha^i, Q_\beta^j \} = 0$, $\{ \overline{Q}_{\dot{\alpha}}^i, \overline{Q}_{\dot{\beta}}^j\} =0$,

$\{ Q_\alpha^i, \overline{Q}_{\dot{\beta}}^j \} = 2 \sigma^\mu_{\alpha \dot{\beta}} P_\mu$

where $P_\mu$ is the 4-momentum operator and the $\sigma^\mu_{\alpha \dot{\beta}}$ are the matrices giving the isomorphism between the vector representation and the tensor product of the left and right spinor representations of the Lorentz algebra ($\sigma^\mu = (id, - \sigma_i)$ with $\sigma_i$ the Pauli matrices). For a massless representation of the supersymmetry algebra, one can assume $P = (E,0,0,-E)$. In this case, the 2 by 2 matrix $2 \sigma^\mu_{\alpha \dot{\beta}} P_\mu$ has only one non-zero entry equal to $2E$. This implies that the supersymmetry algebra acting on this representation has $N$ fermionic creation operators. As a fermionic creation operator changes the spin of a state by 1/2, this means that the range of spin in a massless representation is equal to $N/2$. If $N>4$ then $N/2 >2$ so the representation contains a spin outside the interval [-1,1] : if all massless particles in the theory have spin $\leq 1$ then $N \leq 4$. If $N>8$ then $N/2 >4$ so the representation contains a spin outside the interval [-2,2]: if all massless particles in the theory have spin $\leq 2$ then $N \leq 8$.