Yes, your opamp circuit presents resistance of -R3 to ground at the positive opamp input. I think your question is more about what will a voltage source do when presented with a negative resistance.
The current will flow backwards thru the voltage source. That means the negative resistance is producing power and the voltage source dissipating it. That's OK. In fact negative resistances always produce power unless the voltage across them is zero, just like positive resistances always absorb power unless the voltage across them is zero. This is one reason why we don't have negative resistors like we do positive resistors. The negative resistors would have to produce power.
To see who was paying attention when he was talking about negative resistances, my circuits professor in college ended by saying "... and I have a jar of them in my office. Anyone that wants to see one can come by later." and then looked around to see who laughed. I was surprised how many were staring blankly wondering why a few of us were chuckling.
I think the confusion comes from the fact that we rarely run into negative resistances, and that you are used to thinking of a voltage source as producing power, not absorbing it. If so, you need to broaden your outlook. The only thing an ideal voltage source is guaranteed to do is to hold the voltage across it constant. That is true whether is has to source or sink current to do it. There are instances in regular circuits where we have voltage sources that are intended to work by sinking current. That is basically what a shunt regulator is. A Zener diode is a passive component that does this.
You have to realize that a voltage source, particularly an ideal voltage source used for theoretical analysis as above, is not the same as a power supply. Power supplies may strive to emulate an ideal voltage source, at least for some limits within first quadrant operation, but a true ideal voltage source works for all currents from -ā to +ā.
If the voltage source had a positive resistance, it would be the same as having an ideal voltage source with a positive resistance in series with it. Series resistances add, whether they are negative or positive. The effect from the voltage source point of view is simply the sum of the two resistances.
Note that if the voltage source resistance is too high, then the circuit becomes unstable. Consider a open-circuit negative resistance. It would be stable at exactly 0 volts, but as soon as there was a little voltage on it, it would quickly run away to either positive or negative infinite voltage, depending on the sign of the starting voltage.
Obviously real circuits can't go to infinite voltage. In the case of your opamp circuit, the opamp output can only to its supply rails. Once that happens, it will fail to operate as a negative resistance.
In this case you could just sum R4 and the negative resistance to get the supply current. But if you work out the equations you would get:
\$I_s\$ = \$\frac{\text{R2} \text{ Vg}}{\text{R2} \text{ R4}-\text{R1} \text{ R3}}\$
which is, for the values given is \$I_s\$ = -0.1A , current into the source. This would require a fair amount of voltage out of the OpAmp though.
\$V_o\$ = \$\frac{\text{R3} \text{ Vg} (\text{R1}+\text{R2})}{\text{R1} \text{ R3}-\text{R2} \text{ R4}}\$
For the values given, OpAmp output voltage \$V_o\$ = 20V.
Of course, it all blows up if R4 = Rneg.
Best Answer
You can make such a device using active components. If one end of the resistor is grounded, this simple circuit would work, at least in theory. The op-amp would have to be able to deliver sufficient current and have sufficient voltage swing for whatever voltage you wanted to apply.
simulate this circuit ā Schematic created using CircuitLab
For example, if you apply 5V, the op-amp would have to swing to +10V while sourcing 5A.
An actual negative resistor would be a power source (P = E^2/R) so it's impossible to have such a device that does not use or contain a source of energy.
There are a few devices and circuits that exhibit negative differential resistance- over some region of operation the current drops with increasing voltage (while the total current is always positive). For example, Esaki diodes, discharge tubes, switching power supplies. So in a small signal analysis, resistance can be negative.
Edit:
You've added that you require it to be ungrounded. In that case, merely power it by a battery or an isolated DC-DC converter and use the circuit as shown (with a suitable power op-amp). There are still many "compromises" - as in limitations and non-idealities as there are with any real circuit or element, of course.
Here is a simulation with a 1K resistor for R1 which simulates a -1K resistor.
simulate this circuit
As you can see the current through the simulated -1K resistor is the mirror of the current through the real 1K resistor R4. I've offset the other side of the resistor by 5V from ground in each case and the sawtooth goes from 0 to 10V so the voltage across each element goes from -5V to +5V. Not shown are the power supplies.
You won't easily be able to simulate the isolated version because op-amp models typically have ground nodes within the model.