Negative resistance from MOSFET circuit

Say you increase the voltage at \$V_{DS}\$. This increases the voltage on Q2's gate, increasing the conductivity of its channel, lowering the voltage at \$V_{GS}\$. This decreases the current through Q3's channel. So by increasing the voltage \$V_{DS}\$ you've decreased the current that the \$V_{DS}\$ supply must provide, which is the definition of a negative resistance load.

Of course this all depends on Q1, Q2, and Q3 being in the appropriate operating modes, so there will be a limited range of voltages over which the negative resistance behavior can be observed.

negative current is not proportional to voltage.

If you apply a positive voltage to a circuit branch and it sends current back out towards you ("negative current") then that branch is delivering power to you, not absorbing power from you.

This circuit doesn't do that.

But it does have a region where \$\frac{dV_{DS}}{dI_{DS}}<0\$, which is what the authors are referring to as negative resistance (and also what we call negative resistance in the case of common negative-resistance elements like glow tubes and Esaki diodes).

(I dedicate my answer to @tlfong01 in gratitude for his responsiveness.)

This is a typical example of how a simple idea can be presented in an insanely complex way to become a scientific article...

This kind of negative resistance is named "negative differential resistance" (NDR). Actually, it is "over dynamic resistance" (I have revealed the truth about it in https://en.wikibooks.org/wiki/Circuit_Idea/Negative_Differential_Resistance).

Devices such a tunnel diode possess such an N-shaped IV curve. Here it is artificially obtained. The trick to create the negative resistance region is simple - driving a MOSFET (Q3) both from the side of the drain and gate. Thus, when Vd increases, Q3 is forced (from the side of the gate) to increase enough its dynamic "resistance" so that Id decreases despite the fact that Vd increases.

This can be explained by Ohm's law written in a form of Id = Vd/Rds where Rds is the dynamic "resistance" between the Q3 drain and source. In this expression, when the numerator (Vd) increases, the numerator (Rds) also increases... and to an even greater extent. So, their ratio (Id) decreases.

In other words, the trick is that, in Ohm's law, the current is a function of two variables - both voltage and resistance... and thus its direction of change is reversed.

It only remains to answer the question, "What is the point of all this?"