It is a dynamic equilibrium. To see this notice that the Boltzmann distribution law is statistical in nature, for instance, the molecules that have energies between $E_1$ and $E_1+\Delta E_1$ at time $t_1$ are not necessarily the same than the ones in between those same energies at time $t_2$. Particles collide (and thus exchange kinetic energy) all the time, and while the distribution stays the same, the particles that contribute to different parts of the distribution keep changing.
Update:
It is dynamic, like chemical equilibrium, because the molecules do not all have the same kinetic energy. If, at equilibrium, you randomly pick a molecule, there some probability
(from the Boltzmann distribution) that it will have its energy between $E_1+\Delta E_1$. If after some time $\Delta t$ you pick those same molecules again, there is again the same probability to find those molecules with an energy between $E_1$ and $E_1+\Delta E_1$ is the same, but those same molecules will likely have a different energy than the one you measured earlier. this is because they molecules collide all the time and exchange energy with each other. From a macroscopic point of view the two states, at times $t_1$ and $t_1+\Delta t$, are the same; from a microscopic point of view, the two states are different. Thus it is a macroscopic equilibrium not a microscopic one. Same when have ice and water at equilibrium. The molecules that form part of the ice or part of the water keep changing all the time. At equilibrium, you keep the same amounts of water and ice, but the molecules that make up the ice and the water keep being exchanged between the ice and the water.
I will provide an answer from an astrophysics point of view, in which the term local thermodynamic equilibrium (LTE) is often used. In astrophysics the distinction between 'thermal equilibrium' and 'thermodynamic equilibrium' is not carefully made, because there is rarely if ever a situation in this context in which thermal equilibrium might hold without thermodynamic equilibrium.
The most common situation in which the presence or absence of LTE is considered is for a star. There is a flow of heat from the interior of a star to its atmosphere at large radii, and the temperature varies as a function of radius. So clearly gas near the atmosphere is not in thermodynamic (or thermal) equilibrium with gas in the core.
However, models of stellar interiors can be vastly simplified if one recognizes that there is still a local thermodynamic (thermal) equilibrium in the sense that the kinetic distribution of the free electrons, the plasma ionization state, and the radiation field at each radius can all be very well described by a single number, a local temperature $T$. The electron velocities obey a Maxwell-Boltzmann distribution, the ionization state follows from the Saha equation, and the radiation field is described by a Planck function (blackbody), all evaluated for some common $T$.
As the link you provide explains, this works as long as the mean free path of any particles that might transport heat (e.g. photons, electrons) is very small compared to the length scale over which the temperature is changing. In the atmospheres of stars, where the photon mean free path grows large, LTE in the above sense can break down. However, if the electron mean free path is still small enough, it can helpful to apply an even more limited notion of LTE in which one takes the electron velocities to be in LTE, while acknowledging that the radiation field may depart from a Planck function.
To summarize, LTE is a useful concept to the extent that it simplifies the description of a large system as a collection of local regions each described by a single temperature. The distinction between thermal and thermodynamic equilibrium is rarely needed in the situations in which LTE is a helpful approximation.
Best Answer
This is a question of meaning of words - semantics. I did not make any research into how the terms are used in the classic physics literature, but I think the following view is quite reasonable:
system is in thermal equilibrium when any two parts of it are in direct or indirect thermal contact (capable of exchanging heat) but no appreciable heat is exchanged from one to another even after a long time (concern with heat flow)
system is in thermodynamic equilibrium when all macroscopic processes in it subsided(concern with any macroscopic flow, not only heat).
Yes, for example two gases of different temperature trapped in a cylindrical vessel separated by movable but thermally insulating piston. If the pressure of the gas is the same in both compartments, the system will remain without change indefinitely, so it is in thermodynamic equilibrium (under given constraint of perfect insulation). However, it is clearly not in thermal equilibrium, because there is no possibility of thermal exchange between the two parts.
Admittedly, this is an idealized situation, as there are no perfect thermal insulators in the real world. In practice such a system would not be in thermodynamic equilibrium either, as some heat would flow from the hotter to the colder gas. But it is possible to make walls that would maintain the state of different temperatures for a very long time without appreciable change in the macroscopic state, so this would be a system in thermodynamic equilibrium for all practical purposes.
Yes, for example a gas that is being compressed by a moving wall. The temperature and pressure increases everywhere as work is being performed on the gas elements; but there is no heat flux anywhere. The system is in thermal equilibrium, but not in thermodynamic equilibrium, as macroscopic changes are happening.
Of course, in practice the compression would not occur uniformly in the whole space, but there would be compression waves and heat exchange between elements of the gas. The system is thus not even in thermal equilibrium. But if the compression is done slowly enough, heat exchanges will be negligible and most energy transfer will be via work. Thus the system would be in thermal equilibrium for all practical purposes.