Well the condition will be different for each component. Resonance doesn't matter. The current must obviously be identical in all three series elements. The Voltage across the resistance will be in phase with that current, and the Voltage across the capacitance must lag the current by 90 degrees, while the Voltage across the inductance must lead the current by 90 degrees. This makes the inductance and capacitance Voltages 180 degrees out of phase. At resonance they will be equal and opposite, so they cancel leaving just the whole source Voltage across the resistance.
What you could say is "if energy is lost in a resistor, then why doesn't the velocity of the charged particles increase, as per Work-Energy theorem?". So your question should really go something like "Why doesnt current which is $Q/t$ increase if the velocity will increase after voltage drops"?
The answer to this is that we assume all potential energy lost, is again lost due to inner collisions with other atoms, and that's why materials heat up! Also, this implies a steady current because the drift velocity will not be changing.
Edit:
It can be shown that $F\Delta x= \Delta KE$ this means that a force acting on some distance will produce a change in kinetic energy. You can imagine an electric field where the work done by it is simply $W= qE\Delta x$. By the way notice that if I were to divide that expression by the charge $q$ I would obtain the voltage across the thing i'm concerned with. Namely $V=E\Delta x$.
So in summary, if I do some work, then I have some change in kinetic energy. If there is a voltage drop, it follows that positive work was done and kinetic energy increased, which means a velocity increment.
Across a resistor there is a voltage drop. So why isn't it that charged particles are going faster and then I can measure a current increase? Well that's due to the explanation I gave above my edit portion. Namely, that all that increase in kinetic energy is absorbed due to collisions with the neighboring atoms.
By the way, if you're curious enough to visit this website, I suggest you look up a video on Work-Energy theorem on youtube. This concept is pretty straightforward and I'm sure you can understand it.
Best Answer
No, a phase difference of 180º between applied voltage and current is not possible in a standard LCR circuit.
However, for reference, it is possible to construct electronic circuits that mimic negative resistance or negative impedance. In the case of negative resistance, $R = \dfrac{-V}{I}$. Here is a wikipedia article about it.
This will, for a voltage sine wave produce a 180º phase shift in the current. These types of circuit can use inductors and capacitors and without going through all the possibilities there is probably a way of adding L, C and Rs around an op-amp to achieve it too.