Okay, at last I managed to get some time to prove that circuit #19, with some common sense, needs not overdrive the crystal at all.
The idea is to add a couple of Schottky diodes to limit the drive to the crystal.
For a mathematical proof, the diodes will limit the drive into the crystal to a 200mVpp square wave. If we couple that drive to the first mode of the oscillator, we get the steady state excitation of the crystal. In voltage terms, it amounts to \$\frac{2}{\pi}V_{e}\frac{R_l}{R_s + R_l}\$, where Ve is the square wave excitation, Rs the series resistance of the crystal, and Rl the load resistance. In our case, the load resistance equals the series resistance, so the sinusoidal amplitude in the crystal reduces to \$V_e/\pi\$. The power dissipation is \$P \ = \ \frac{V_e^2}{2\pi^2R_s}\$ which, taking Ve=200mV and Rs=33775 Ohm, gives P = 0.33uW. Not bad. LTSpice is more optimistic and predicts about 0.2uW.
I've built a quick version of this circuit, and it works, but with quite a few parasitics. I think the problem is the layout; if the diodes are removed the parasitics remain. In fact, the circuit self oscillates without crystal.
In the picture I'm using a 100kHz crystal. I've also tested the circuit with a 32kHz crystal. I attach traces for the 100kHz oscillation, the 32kHz oscillation, and a trace of the schottky section at 32kHz. The RMS square wave is 163mV RMS, so the drive to the crystal is very soft.
I can improve the oscillator to make it cleaner, but think this goes a long way proving that the two inverter topology needs not overload the crystal if done right.
Sorry about the delay answering this. I've have had a long, hard week.