What is the resolution of this Lewis-Tolman-like paradox? Though my question stands on its own, here is a brief overview of a well-explored question, the Lewis-Tolman Lever Paradox, which I think might be helpful in finding the resolution to my question. The Lewis-Tolman Lever Paradox has been stated as follows :

When a rigid right-angled lever experiences equal and opposite torques as measured by an observer O′ at rest with respect to the lever in frame S′, the usual transformation of these torques to another Lorentz frame S makes it appear that the torques are not balanced for an observer O at rest in S.... The consensus among textbook authors is that a net torque does exist on the lever according to the general Lorentz observer.
Reference: http://scitation.aip.org/content/aapt/journal/ajp/38/3/10.1119/1.1976326

Consider an elastic rod attached to a small sphere of charge $q$ at one of its endpoints and to an identical sphere at the other end. In a frame $O$, the entire system is at rest and in equilibrium. The system lies in the $XY$ plane and the axis of the rod makes an angle $\theta$ with the positive direction of the $x$ axis. In this situation, the force on one of the spheres is $\textbf{F} =$ $F \cos\theta$ $\hat{i}$ $+ F\sin \theta$ $\hat{j}$ and that on the other is $-\textbf{F}$. Both of the forces, of course, pass through the center of the corresponding spheres and also lie along the axis of the rod. The length of the rod is $l$. The endpoints of the rod can be identified as $(0,0)$ and $(l\cos\theta,l\sin\theta)$ respectively. Notice that this is the length in the equilibrium position in the rest frame $O$. It is clear that an analysis done in the frame of $O$ suggests that the rod should be stretched but not bent.
Now I choose a frame of reference $O'$ with its spatial axes being parallel to those of $O$ as observed at any instant by $O$. $O'$ is moving along the positive direction of the $x$ axis with a speed $v$ as observed by $O$. The rod, as observed by $O'$ at any particular instant, will be a rod with its end-points at $(-vt',0)$ and $(-vt'+l\sqrt{1-v^2}\cos\theta, l\sin\theta)$, i.e., the axis of the rod makes has a slope $\dfrac{1}{\sqrt{1-v^2}} \tan\theta$.
Now, the force transformation laws between two inertial frame suggest
$F_{y'} = (F \sin \theta ) \sqrt{1-v^2}$ and $F_{x'} = F \cos\theta $ , i.e., the direction of the force makes a slope $\tan\theta \sqrt{1-v^2}$.
Thich means that the forces should bend the rod as judged by $O'$. Because at every instant the line of action of the forces does not coincide with the axis of the rod. This contradicts the judgement reached by an analysis done in frame $O$. Thus, the paradox.
 A: This paradox, like the Lewis-Tolman paradox, is basically the result of trying to use three-vectors in a four-dimensional theory.
The four-force points in "the direction that the worldline is pushed", and is always perpendicular to the worldline. That means that its $t$ component in frame $O$ is zero, so $\mathbf F = F \cos θ\,\hat x + F\sin θ\,\hat y$ still. After the Lorentz boost, you get
$$\mathbf F = -γβ\,F\cos θ\,\hat{t'} + γ\,F \cos θ\,\hat{x'} + F\sin θ\,\hat{y'}$$
If you draw a line with a direction of $\mathbf F$ through any point on the worldline of one end of the rod (such as the origin), you'll find that it intersects the worldline of the other end. So the force is still in line with the rod. Of course, this is a foregone conclusion, since a mere change of coordinates can't change the geometrical relationships between the objects in the problem.
If you drop the $t'$ component of the transformed force, in effect you're adding a scalar multiple of $\hat{t'}$ to it. That vector isn't aligned with anything else in the problem, so the force points in the wrong direction after you add it.
It's possible to define three-vector versions of many four-dimensional quantities, but they're only meaningful because they can be transformed back into four-dimensional quantities. In most cases that involves adding a $t$ component which changes their direction. This paradox would probably be resolved in three-vector terms by saying that the acceleration from a force doesn't point in the same direction as the force, and a complicated calculation would show that everything works out. It's easier to just use four-vectors.
A: It has been a long time since I posted the question so I am not sure what was the motive behind not following through the paper I already linked in the question. ;)
As pointed out in the linked paper, a relativistically covariant definition of the torque differs from the Newtonian definition of the torque. If such a definition of the torque is applied, all Lewis-Tolman-like paradoxes are doomed to be evaporated. And indeed, that's what happens to this example.
As elaborated in the paper, the relativistically covariant formulation of the torque ought to be done as a tensor arising out of the commutator of the four-force vector and the four-displacement vector. If one then takes the three-space part of this covariant object then they can choose to represent it as an axial vector as usual but the expression for it in terms of the lengths and three-forces differs from the Newtonian expression. Namely, we have
$$\vec{\tau}=\pmatrix{\hat{i}&\hat{j}&\hat{k}\\\gamma^2l_x&l_y&l_z\\\gamma F_x&\gamma F_y&\gamma F_z}$$
In this example, this means that the torque produced by the present pair of forces is given by
\begin{align}
\vec{\tau}&=\hat{k}\Big[\big(\gamma^2l\sqrt{1-v^2}\cos\theta\big)\cdot\big(\gamma F\sin\theta\sqrt{1-v^2}\big)-(l\sin\theta)\cdot(\gamma F\sin\theta)\Big]\\ &=\gamma lF\sin\theta\cos\theta - \gamma lF\sin\theta\cos\theta\\
&=0
\end{align}
Thus, even if the three-forces are not aligned along the axis of the rod, the torque produced by them is still zero and thus, there would be no bending of the rod -- in agreement with the physical conclusion reached from the rest frame.
Notice that while, in general, a net-zero torque does not imply the lack of bending, in this case, the calculation shown above does imply a lack of bending because it shows that the torque by each of the forces is zero individually. And thus, the rod would not need to bend.
