How to avoid Johnson noise in high input impedance amplifier
The problem in your reasoning is that you do not show the complete path of the signal. More specific the impedance level of the signal.
You are right in that you cannot have both a high impedance and a low noise. If you want low noise you must keep the impedance low. Simple as that.
In the two circuits you have drawn it is unclear what the impedance is of the source which you use to feed a signal to your amplifier. Assuming that the AC coupling capacitors are large and that this source impedance is low (for example: 50 ohms) then the noise will be low !
Why ? Because the noise generated by the 100 Mohm DC bias resistors will be shorted by the AC coupling capacitors and that low source impedance. So in this situation the effective signal impedance (at a certain frequency) is much lower than 100 Mohm. Resulting in a low noise.
If the 50 ohms source impedance was not there that noise current would multiply by the 100 Mohm of the resistor itself resulting in a high noise level.
You can do calculations on this more easily by considering the noise current generated by the 100 Mohm resistors. That current will be multiplied by the signal source impedance (for example, 50 ohms) resulting in a small noise voltage !
So the circuit on the right is no better than your circuit on the left. Carefully read how they measured that low noise and try to figure out what the impedance level of the input signal was. I guarantee you that they will have used a source impedance such that the noise of the 100 Mohm DC biasing resistors can be neglected (a very low source impedance, they might even have shorted/grounded the inputs !). In that circuit the noise of the FETs should be dominant as these should determine the lowest possible noise level (at least in a properly designed amplifier).
Remember that you are connecting this amplifier to a signal source, so this 200M impedance is in parallel with the source impedance.
Measure the amplifier's noise with the input open circuit and you will see your predicted noise. (plus a contribution from any electric fields at the input; you may need screening to measure this properly)
Measure the amplifier's noise with the input short circuited and you'll see the amplifier's own inherent noise.
Measure the amplifier's noise with the actual source impedance it'll be connected to and you'll see the amplifier's own inherent noise. The ratio of this to the noise of the source impedance alone is the amplifier's "noise figure".
With a 10Megohm source resistance (leg to leg) you'll see Johnson noise from 2 resistors in parallel - 10Meg and 200 Meg, so you may see 0.5dB less noise than a 10Meg resistor alone (but you have attenuated the signal by the same fraction too)
With a capacitive source - such as a 30pf microphone capsule, the source impedance is a parallel R-C network, so treat the Johnson noise as the noise voltage from 200M, attenuated by a 200M source impedance into a 30pF capacitor. It'll be nominally flat up to the -3dB frequency, then reduce by 6dB/octave.