In the same way that a diode lowers the voltage typically by 0.6V/0.7V, I would like to understand the minimum and simplest circuit to obtain the reverse, i.e. raise the voltage by 0.6V/0.7V compared to the input voltage.
We are talking about low and constant input voltages over time, typically from 12V to 13V
I know about stepup circuits, but the question is whether there is something much simpler.
A diode drops voltage and the lost power (volts dropped multiplied by current) is dissipated as heat.
If you want to increase the voltage you need to supply power from somewhere so it cannot be a 2 terminal device unless that device includes a power source such as a battery, thermopile, PV cell or something like that. That's just from conservation of energy.
A PV output optoisolator operating into a diode (two components plus a series resistor for the optoisolator IR LED) is the simplest circuit I can think of offhand, and it would be capable of maybe 5 or 10uA.
There may well be a very simple approach but it's really very much situation dependent. If you want something that involves significant power, there is no higher voltage source available nor a voltage source that can be put in series, then a boost converter is probably the simplest solution. A boost converter at its core is just an inductor, a capacitor and a couple switches (the latter typically in an IC these days). The output voltage is not a fixed adder but rather a ratio of the input voltage (ideally 1/(1-D) where D is the duty cycle of the switches). Usually IC boost converters involve feedback and a controller, which you could use to adjust D to get a fixed voltage increase with some circuitry.
On the other hand if you just have a signal and want to add some voltage to it (and have a higher voltage power supply available), there are various series diode with bias resistor or op-amp circuits that might be suitable, again very much dependent on the details of the application.
The reason a diode is able to "lose" 0.7V, is because it is taking energy away from charges. Specifically, it is removing 0.7 electron-volts (0.7eV) of potential energy, and dispensing with that energy as heat.
If you wanted to raise the potential by 0.7V , then you have to add 0.7eV of potential energy to each electron, which requires a source for that energy. A source whose potential is already at least 12.7V.
In an analogy, you can gently nudge a marble off a 0.7m high table without lifting the marble. If you wanted to return the marble from the floor to the table though, you have to lift it, and the only way to do that is by expending your own energy.
Given that you need a source of potential greater than 12.7V, here's how you might achieve the "0.7V lift":
simulate this circuit – Schematic created using CircuitLab
Here the potential energy of charges traversing the diode changes by 0.7V (like a 0.7m "fall"), which is why the potential at the top of the diode is 12V + 0.7V. Often, thinking in terms of energy like this, helps to better understand what's going on.
DC to DC converters use electrical gymnastics to figuratively "throw" charges to a higher potential, in spite of their tendency to always fall in an electric field, to a place of lower potential energy. In this way they are able to produce potentials greater than their own energy source.
That doesn't mean you get free energy, it just means that you get a few charges with, say, 24eV potential energy, energy donated and lost by twice as many charges with 12eV of energy.
In answer to your question "is there a simpler way?", conservation of energy is law, and if you want greater potential somewhere, then you have to give the charges there extra potential energy. That requires a source of energy somewhere else. The simplest way is in my schematic above, but I fear it's not the answer you wanted.
