It did look odd as I typed it and had to retrace my steps to convince myself that I hadn't overlooked something. Consider three identical resistive loads connected in star to form a perfectly balanced 3-phase 3-wire load. Everything is perfectly insulated from earth, so cannot produce any earth leakage / RCD differential current. The three currents at the load star point vectorially sum to zero and the star point will take up the same voltage as the supply neutral. Therefore it doesn't matter whether the star point is connected to neutral or not; no current will flow along the neutral conductor because there's no potential difference between the load star point and the supply star point.
If protective earth and neutral are at the same potential, then the same is true of a connection between the star point and earth. The star point could be solidly connected to the system CPC without producing earth-leakage. If we consider that the CPC has zero resistance, the three single-phase loads could be separately connected to their own CPCs without creating differential current upstream of their interconnection point, because although there are now currents circulating through the CPC they still sum to zero.
With ideal components there's no limit to the total symmetrical leakage that could be accommodated, but in the real world there are so many sources of asymmetry that one probably can't get away with very much. One should perhaps assume that only a total leakage less than I-delta-N is acceptable for practical use, but might nonetheless find that exceeding this limit doesn't always cause a trip.
Also worth noting on a different subject that for real loads such as transformers and motors, the internal star point is generally not connected to neutral to prevent circulating currents due to asymmetry; it is not true in practical applications that zero potential exists between the supply and load star points.