I don't know of any intuitive argument. It may have to do with the fact that the irreps of higher spin are necessarily "gauge fields". The reason is that the little group for the massless representations is ISO(2), i.e., the symmetry space of 2D Euclidean affine space. The little group is the subgroup of the Poincare group which leaves one arbitrarily choosen four-momentum, which for the massless representations is light-like, invariant.
Now you need the unitary transformations of the little group to build the irrep. of the full orthochronous proper Poincare group a la Frobenius/Wigner. The ISO(2) is generated by 2 translations and 1 rotation. The corresponding generators for the translations have of course continuous a continuous spectrum. Realizing this would mean you'd describe massless particles with continuous polarization-like degrees of freedom, which is physically not observed. Thus you consider only such representations, for which the translations of this ISO(2) are represented trivially. This leaves only the rotations of the ISO(2) as non-trivial (the trivial one leads to scalar massless fields though). From the point of view of the entire Poincare group these are the rotations with the rotation axis around the three-momentum defined by the arbitrarily chosen light-like four-momentum, i.e., the generator of these rotations is the helicity, and together with the triviality of the translations of the said little group this implies that you'd have all possible helicities ##h \in \mathbb{R}##, but at the end you use the Frobenius-Wigner method to build the representation of the entire Poincare group including the usual rotations as a subgroup, and this means that the helicities can only be ##0## (scalar fields), ##\pm 1/2##, ##\pm 1##, etc. leading to the representations corresponding to spin ##s=0##, ##s=1/2##, etc. for the rotation group as a subgroup of the Poincare group. So for any non-zero spin of your massless field you get only 2 polarization-degrees of freedom instead of ##2s+1##. For spin 1/2 you get both ##\pm 1/2## for each spin component. For ##s \geq 1## you get more spin-like degrees of freedom than the two physical polarization referring to the helicities ##\pm s## for each energy-momentum eigenmode, i.e., you have to deal with unphysical degrees of freedom, making your theory a "gauge theory" when expressed in terms of local fields.
That's why for the electromagnetic field for each energy-momentum eigenmode you have only the two transverse polarization states corresponding two left- and right-circular polarization (i.e., ##h=\pm 1## helicities) and not 3. Of course you want to work with local fields (in both classical and quantum electrodynamics). That's why you introduce a four-vector ##A^{\mu}##, but all fields connected by a "gauge transformation" ##A^{\prime \mu}=A^{\mu} +\partial^{\mu} \chi## with an arbitrary scalar field ##\chi## are considered to describe the very same physical situation. Only such a description with redundant, unphysical field degrees of freedom allow for a local description, and that's why we have to deal with the problems connected with gauge invariance when quantizing the theory. One way is to fix the gauge completely (e.g., by choosing the Coulomb gauge) and loosing manifest covariance (leading nevertheless to a covariant field theory as far as the physical fields and the observables built from them are concerned) of to use the tricky Faddeev-Popov quantization by introducing more unphysical ghost fields which precisely cancel the contributions from the also unphysical parts of the vector potentials. This modern path-integral method is equivalent to the manifestly covariant operator method. For the Abelian gauge-theory case of QED that's equivalent to Gupta-Bleuler quantization, but for the non-Abelian case you need the Faddeev-Popov ghosts for a gauge-fixed manifestly covariant formulation and the associate Becchi-Rouet-Stora-Tyutin (BRST) symmetry of the gauge-fixed action. Here, I'd say, the original path-integral formulation aka Fadeev-Popov quantization is more straight-forward.
For massive particles the little group is simply SO(3), and there you have all ##(2s+1)## spin eigenvalues as physical states.
Maybe it's this difference between the massless and massive irreps. which allow for the development of some intuitive argument for the non-existence of proper position observables in the massless case for spin ##s \geq 1##.