Why is the number of thermodynamic systems (that do not do work) in this scenario two? The hardest thing for me about studying thermodynamics, is connecting the formal theory with actual problem instances. Here is an example:
In Callen’s Thermodynamics and an Introduction to Thermostatics (second edition), Chapter 4, Example 1, the following scenario is described:

A particular system is constrained to constant mole number and volume, so that no work can be done on or by the system. [...]. Two such systems, with equal heat capacities, have initial temperatures $T_{10}$ and $T_{20}$, with $T_{10} < T_{20}$. An engine is to be designed to lift an elevator (i.e., to deliver work to a purely mechanical system), drawing energy from the two thermodynamic systems.
What is the maximum work that can be so delivered?

What I don't understand here is: Why does the question mention two such systems? More generally, I don't understand how I'm supposed to envision this scenario.
I understand that the given system cannot do mechanical work, so it can only transfer (or receive) heat to another system, which I believe is the engine. Therefore, the engine needs to be able to do the mechanical work. What I imagine is: the given system is in contact with an engine. The engine is simply (for example) a gas in a cylinder with a movable piston. The gas in the engine is heated by the system, and therefore expands, and this moves a piston which gives us the required mechanical motion. I understand that this process can continue until the temperature of the given system equals the temperature of the gas in the engine.
So where does the second system enter? Why do we need it? Am I imagining the scenario correctly?

EDIT
Here is a synthesis of all the answers I received, in words that are easier for me to understand. I hope I got it right.
We are supposed to think about the engine as a rather limited entity, in at least these aspects:

*

*It has a small heat capacity. This means that, when it comes into contact with the high temperature system (which I'll call system $A$), a small amount of heat transferred to the engine is enough to raise its temperature to the level of system $A$. At this point, no more energy (heat) can be transferred to the engine from system $A$.


*The engine's ability to do work might be mechanically bounded. For example, if the work is performed by expanding gas in a piston, the piston might have a maximal expansion length, after which it cannot do anymore work (because the piston cannot mechanically move anymore).
It follows that, when running the process, the engine will "quickly" get from its initial state to a final state in which it cannot do anymore work - either because it cannot take in anymore heat from system $A$ (this heat is what drives the work to be done), or because it is mechanically "stuck".
However, more work can be done if the engine is brought back to its initial state, and then the entire process is repeated. But to bring the engine back to its initial state, the engine must dump heat to another system - a low temperature system (which I'll call system $B$) - so that the engine's temperature can return to the initial value. System $B$ allows the engine to work in cycles, and hence to perform more and more work.
As these cycles are repeated many times, system $A$'s temperature is slowly reduced, and system $B$'s temperature is slowly increased, until both temperature become equal. At this point, no more work can be done.
If the given scenario did not include system $B$, then we would need to consider more carefully the details of the engine, e.g its temperature and heat capacity. In this case, the engine would take heat from system $A$ until the temperatures are equal, while performing work. And we don't have repeating cycles.
 A: 
What I don't understand here is - why does the question mention two
such systems? More generally, I don't understand how I'm supposed to
envision this scenario.

Simply think of the two systems as a heat source and heat sink between which a heat engine can operate. Heat is taken from the high temperature system, partially converted to work, with the remaining energy transferred to the cold temperature system per the second law, until the two systems reach thermal equilibrium.
The maximum amount of work that can then be obtained using the two systems is that produced by a Carnot heat engine cycle operating between the systems. But since the systems are not thermal reservoirs, an infinite number of Carnot cycles need to be carried out between the two systems, with each subsequent cycle producing less work than the prior cycle,  until the two systems reach the same temperature and no further work can be obtained.

why isn't one system enough? why do we need that "the remaining energy
transferred to the cold temperature system" ? why can't the engine
take heat from the high temperature system, convert some of it to
work, and that is that ?

That would violate the Kelvin-Planck statement of the second law, which is:
No heat engine can operate in a cycle while transferring energy with a single heat reservoir
Hope this helps.
A: There are two factors that haven’t been mentioned yet in the existing (very good) answers: one is an implicit assumption, and one imposes in conjunction a fundamental necessity for two systems/thermal reservoirs (as opposed to only one such system).

*

*Callen’s engine is implicitly assumed to not contain an additional hot or cold reservoir. That is, it is implicitly assumed to have negligible heat capacity. Why? Because it's stated elsewhere in the problem that a net entropy change is allowed in the reservoirs only—not the engine. (An alternate assumption, already mentioned, is that the engine is simply returned in the same state it was provided. In that case, for this particular problem with a final temperature of $T_f=\sqrt{T_{10}T_{20}}$ for both reservoirs, calculated here, any heat engine with a non-negligible heat capacity would require that exact initial temperature at the beginning. It's simpler to assume a negligible heat capacity.)This is why the responses tend to discuss cyclic operation of the engine: It’s the only reasonable way to extract the maximum work, as tasked. The reader is expected to realize that the engine must collect some thermal energy, output work (among other required actions, specified below), collect some thermal energy, output work, etc. In other words, it cannot reasonably consume the available energy content of the hot reservoir in a single step, as it itself doesn’t have the capacity.If one cycle extracts work $W$, then $n$ cycles extract up to $nW$ total work—depending on how large the reservoirs are—which is better. Of course, finite thermal reservoirs will be gradually depleted with continued operation; they themselves are not cyclically regenerated. The final state of a common temperature between the two finite reservoirs is asymptotically approached in the ideal case as the number of cycles increases.
(Is this a frustrating mix of idealizations—minimal temperature gradients to avoid generating entropy, for example, and a negligible heat capacity—and the reality that an actual heat engine must have a finite size and be driven by finite temperature gradients? Yes; and this is common to thermodynamics thought experiments involving thermodynamic reversibility and Carnot engines. The Carnot engine can't be built—all real processes are thermodynamically irreversible—yet we discuss it constantly. Physical approximations to it would have a power output of essentially zero. When Callen refers to the "maximum work" but a "common [final] temperature," the thermodynamics practitioner must immediately conceive of a quite unusual combination of very large hot and cold reservoirs connected by a very small heat engine, transferring minuscule heat and work very slowly over an arbitrarily large accumulation of cycles. Why go to this cognitive effort in assembling an improbable tower of assumptions, idealizations, and impractical assemblies? The payoff is not having to incorporate entropy generation calculations; entropy transfer calculations through heating and cooling are sophisticated enough at this stage.) 


*Work doesn’t transfer entropy, but heat does, and entropy can’t be destroyed. This is the fundamental constraint of heat engines.  Callen takes care to note that the reservoirs can’t provide work or matter—they can only heat or cool other things. This means that any energy obtained from the hot reservoir brings along entropy, and since entropy can’t be accumulated in cyclic engine operation, it must be dumped in a second location: the cold reservoir. This requirement holds for all heat engines (and does not constrain other engine types).
These two aspects seem to directly address many of your follow-up comments/questions.
(Edit.) A final note: I think you may be trying to ask the question Can a single thermal reservoir produce work in a single step? Yes, if a gradient other than a temperature gradient is available. For example, abandoning Callen's restriction of constant reservoir volume, you can let the reservoir expand slowly into a surrounding vacuum forever, until its volume is arbitrarily large, its temperature is arbitrarily low, and its thermal energy has been converted essentially entirely into work. This specific example is a pressure-difference engine, not a heat engine. And so the Second Law doesn't prohibit this operation, as the entropy increase from expansion counteracts the entropy decrease from cooling. But you can only do it once; external energy is required for recompression. And it's not what Callen aimed to teach in this example problem, although it's independently an instructive thermodynamics thought problem.
A: I assume a heat engine cycle operating in a cycle. The two systems are a source and sink for heat, respectively, for a heat engine. Based on the second law of thermodynamics, heat cannot be converted completely into work; therefore, not all the source heat can be converted into work and some heat must be rejected.
One of the thermodynamics textbooks by Sonntag and Van Wylen provides a clear discussion of heat engines related to the second law.
A: 
More generally, I don't understand how I'm supposed to envision this scenario.

I want to focus on this part.
The scenario as described mashes abstract and specific in a very weird way, so I would argue that it cannot really be envisioned at all. (Compare it to envisioning: “An object with momentum $p$ hits John Cleese inelastically.”) So, here are two ways of envisioning it:
The abstract way (as in thermodynamic textbooks)

*

*The high-temperature system transfers heat to a piston, being the main part of your engine.

*The gas in the piston expands, doing some work.

*The piston transfers heat to the low-temperature system thereby contracting. This may require some work.

*Go to Step 1 until the high- and low-temperature systems have thermally equilibrated.

A crucial point here is that the piston starts at some temperature, has a finite heat capacity, and maximum expansion length. Therefore, in Step 2, you cannot just transfer an arbitrary amount of heat and convert it to work. Also, you usually want to return the engine to its original state at the end of the process – which requires a way to cool down the engine.
This is why you need the second system. Otherwise your piston would expand once, leaving it in a useless state, and that’s it.
(All of this isn’t mentioned in the quoted scenario, which makes the maximum work infinite, e.g., if you have an infinite piston that is initially filled with an ideal gas at 0 K and does not need to be returned to its original state.)
The practical way
Let’s take the example of a combustion engine:

*

*The high-temperature system is not a thermodynamic system at all but chemical energy stored in a fuel mixture.

*The low-temperature system is the surrounding.


*

*A gas-mixture in a piston is ignited. This transfers energy from the high-temperature “system” to the piston.

*The gas in the piston expands, doing some work.

*The waste gas is exhausted into the surrounding (heating it up) and exchanged with fresh fuel mixture. This requires some work that is either stored in the engine (e.g., as angular momentum) or provided by a second piston operating in a different cycle.

*Go to Step 1.

This may seem like a completely different thing, yet for the thermodynamic budget and efficiency considerations, it makes little difference if you replace Step 3 with something like adiabatic cooling.
