What does negative impedance mean in the context of miller's theorem
The circuit you have is not entirely accurate.  The voltage sources must be connected to ground as the node voltages must be defined with respect to ground for Miller's theorem to work.  Also, there is usually some well-defined relationship between the voltages.  For example, the nodes might be connected with an amplifier with gain K that sets the voltage of one node based on the voltage of the other node.  It would be more accurage to replace V2 with a dependent source that is 3/2 V1.  If K is not constant, then the impedances will depend on the ratio of the actual voltages, which likely won't help simplify a circuit very much.  

As for the impedances, they aren't representative of physical impedances.  It's not possible to make a resistor with negative resistance.  It is, however, possible to make a circuit that acts like a negative resistance in a limited capacity.  In this case, since the 2V node is connected to a node at a higher voltage via a resistor, current will be flowing into the node through the resistor.  When that connection is redefined as an impednace to ground, it must be a negative impedance otherwise the current would be flowing the wrong way.  Check this by calculating what the current would be through the original impedance and for the transformed impedances.  

These sorts of theorems are mathematical tools to help simplify larger circuits so that they are easier to characterize mathematically.  Unfortunately, that means that the implications can be rather abstract as sometimes you end up with seemingly nonphysical results like negative impedances.