I've also no clue, why the quantum issue came up at all. I've also no clue, why this is called a paradox, but maybe that's because I've learned classical electrodynamics and optics quite a time ago already.
As
@Dale stresses, all this is completely understandable with classical electrodynamics and there's nothing mysterious nor paradoxical.
Without going into the microscopic details just consider the most simple polarizer, a polaroid foil, is a device which let's an electromagnetic wave polarized in one direction (say the ##x##-direction) completely through and absorbs one polarized in the direction perpendicular to it (say the ##y## direction, and take the ##z## direction as the direction the wave travels, i.e., the direction of ##\vec{k}##).
Now an arbitrary electromagnetic wave with ##\vec{k}=k \vec{e}_z## can be written as
$$\vec{E}(t,\vec{x})=(E_{0x} \vec{e}_x + E_{0y} \vec{e}_y) \exp(\mathrm{i} k z -\mathrm{i} \omega t),$$
where the physical field is of course just the real part, but the complex notation makes life much easier particularly in this polarization business.
So we can describe the polarizer oriented such that it let's ##x##-polarized waves through and absorbs ##y##-polarized waves completely by the linear (projection) operator.
$$\vec{E}'(t,\vec{x})=\hat{P}_x \vec{E}(t,\vec{x})=E_{0x} \vec{e}_x \exp(\ldots).$$
The intensity ##I## is given (up to an uninteresting constant) by
$$I=\langle \vec{E}^* \cdot \vec{E} \rangle_{T}=\vec{E}_0^{*} \vec{E}_0.$$
Now if you have linearly polarized waves we have ##E_{0x},E_{0y} \in \mathbb{R}## and you can write
$$\vec{E}_0=E_0 (\cos \varphi \vec{e}_x + \sin \varphi \vec{e}_y).$$
The intensity (before the polaroid foil) thus is
$$I=E_0^2=E_{0x}^2+E_{0y}^2.$$
and after
$$I'=E_{0x}^2=E_0^2 \cos^2 \varphi,$$
which is Malus's Law for linearly polarized light.
Now it's easy to show that for a polaroid with it's direction oriented with an angle ##\vartheta## to the ##x## axis acts as the projection operator
$$\hat{P}_{\vartheta}= \vec{e}_{\vartheta} \vec{e}_{\vartheta}^{\text{T}}=\begin{pmatrix} \cos \vartheta \\ \sin \vartheta \end{pmatrix} (\cos \vartheta,\sin \vartheta),$$
i.e., for the amplitude after the polaroid you have
$$\vec{E}_0'=\vec{e}_{\vartheta} (\vec{e}_{\vartheta} \cdot \vec{E}_0).$$
Now it's clear that with two filters in perpendicular relative orientation (say one in ##x## and one in ##y## direction) you get
$$\vec{E}_0'=\hat{P}_{0} \hat{P}_{\pi/2} \hat{E}_0 = \vec{e}_x (\vec{e}_x \cdot \vec{e}_y) \vec{e}_y \cdot \vec{E}_0=0.$$
If you now have first one in ##x## then one in ##\pi/4## and then one in ##\pi/2## orientation you have
$$\vec{E}_0' = \hat{P}_{\pi/2} \hat{P}_{\pi/4} \hat{P}_{0} = \vec{e}_y (\vec{e}_y \cdot \vec{e}_{\pi/4}) (\vec{e}_{\pi/4} \cdot \vec{e}_x) E_{0x}=\frac{1}{2} \vec{e}_y E_{0x}.$$
The intensity after the filter thus is
$$I'=\frac{1}{4} |E_{0x}|^2.$$
Where is here a paradox? Where do I need photons?
