This is not a direct answer, but an investigation of an analogous situation that helped me understand Dale's answer. I'm posting it here in case anyone else finds the discussion illustrative. (But you should still upvote Dave's response!)

This question is part of a more general class of phenomena: a spatially-extended system in which a conserved quantity travels in a loop. Examples include:

- a rotating flywheel (conserved quantity: mass),
- an electric circuit (conserved quantity: electric charge), and
- fluid pumped through a loop of pipe (conserved quantity: mass, vorticity, suspended particles — anything transported by the fluid).

It's very easy to analogize to the situation where the conserved quantity is electricity and flows counterclockwise. The nice conceptual feature of this analogous system is that it can give us a clear division into the black-hole-internal and -external portions.

To do this, consider an arbitrary division of the system into two sections. One section $L$ lies on the left; the other section $R$ on the right. We can assume that the division point occurs at position $x$ both on the top and bottom, where it crosses nothing more complicated than a wire. Importantly, however, we *cannot* assume that this division is time-invariant. The event horizon will pass through our system at lightspeed; $x$ must travel with it.

At any given point in time, we can then describe our system in terms of two quantities and four (signed) flows from right-to-left:

- $q_L$, the total charge on the left;
- $q_R$, the total charge on the right;
- $i_{T,F}$, the current along the top at $x$ (holding $x$ constant);
- $i_{T,B}$, the pseudo-current from momentarily fixing the charges and moving the boundary along the top;
- $i_{B,F}$, the current along the bottom at $x$; and
- $i_{B,B}$, the bottom pseudo-current.

Suppose temporarily that $x$ *is* constant. Then $i_{T,B}=i_{B,B}=0$. By the continuity equation, we have $$\frac{dQ_L}{dt}=i_{T,F}+i_{B,F}$$ But we cannot have unbounded charge accumulate on the left, so in the long-term equilibrium, we must have $$i_{T,F}=-i_{B,F}\tag{1}$$ Indeed, we may already assume that the system has reached this equilibrium. Since $i_{T,F}$ and $i_{B,F}$ are defined by holding $x$ constant, (1) must always hold. Since the current flows counterclockwise, each side of (1) is positive.

Now let the system fall into a black hole (at the left) and choose $x$ to always concide with the event horizon. By the equivalence principle, if our system is sufficiently small, this period should be "nothing special".

We can combine $i_{T,F}$ and $i_{T,B}$ to obtain the total charges falling into the black hole at top and bottom (respectively): \begin{align*}
i_T&=i_{T,F}+i_{T,B} \\
i_B&=i_{B,F}+i_{B,B}
\end{align*} But nothing can exit an event horizon, so we must have $i_T\geq0$ and $i_B\geq0$.

This, then, is the heart of the "paradox": our intuition is formed by situations in which $\frac{dx}{dt}$ is small, if not $0$. In that case, $$i_B\approx i_{B,F}<0$$ When we fall into the black hole, a large and positive $i_{B,B}$ must instead dominate.

But since $\frac{dx}{dt}=c$, a large, positive $i_{B,B}$ is not hard to arrange.

This means that your head and your feet cross at the same instant. The entire flywheel, no matter how large, crosses the horizon all at once." I fully admit I was relying too much on intuition of flat spacetime at the time I wrote my question. However, I am skeptical of the claim in the quote. Are you sure? I'm considering a diagram with Eddington-Finklestein coordinates plotting $t_{EF} = v-r$ vs $r$; see page 6 of here. Now if an object of finite length crosses $r=2M$, its ends cross at different $t_{EF}$. $\endgroup$your head and your feet cross at the same instant," doesn't that presuppose a notion of simultaneity? Are we even able to talk about that in this scenario? If so, how? This discussion might warrant a question page of its own, so I may make another question asking for clarifications. $\endgroup$3more comments