Why is there a constant -1A current through the voltage source at resonance?
C1 takes zero current, as you'd expect.

L1 takes an offset current, because of the phase the voltage happened to start.

The DC current through V1 is correctly shown as the sum of these two components.

If you want this DC current to 'settle down' to zero as time goes on, then you need to add a small resistance in series with the L. This will generate a voltage drop due to the current flow, this voltage will steer the DC current down to zero, with a time constant of L/R. At the moment, the L is ideal, and this persistent DC current is simply ideal behaviour. 

The classical mathematical solution to this sort of problem contains both steady state and transient terms. The oscillatory behaviour is steady state. The 1A DC is a transient. Normally the transient would die out with an L/R time constant. In your ideal case with R=0, L/R is infinity, and the transient does not die out.

Delaying the start of the simulation won't necessarily change the start phase of the sinewave from V1. Depending on what simulator you're using, you should be able to change this in the V1 source parameters directly. If you have a cosine option, this should give you zero offset current. Bonus marks for why sine that starts at zero gives you an offset current, and cosine that starts at maximum voltage will give you zero offset current.

Even if you apply a cosine voltage waveform, the transient behaviour still last indefinitely, it's just that the transient has value zero.