Please note there is a difference between power factor (PF) and displacement power factor (DPF)
PF is associated with the harmonic content while DPF is associated with the fundamental
DPF = cos(\$\Theta\$)
PF = \$\frac{DPF}{\sqrt{1+THD^2}}\$
why is this important? Because there are two scenarios where the real power can be zero.
1) Purely reactive load causing 90° phase shift between voltage and current
2) Purely harmonic current draw, causing no current draw at the fundemental.
colloquially, PF is used interchangeably with DPF more freely than it should be. In the scenario you are considering, the PF would equally the DPF as the THD = 0
What would happen? Well assuming you could acquire an ideal inductor and you had ideal cabling to your house, you would draw current equal to \$ \frac{V}{j\omega L} \$ while drawing no real power. This current could potentially be quite high.
Domestic billing is measured and charged in WATTS so you would theoretically not be charged anything. The grid and the generation however still need to source this current and this is where the problems are.
Industry are charged in VA so every harmonic or DPF they draw, they must pay for since the generation and distribution must manage this.
If you or an industry complex did draw excessive DPF or PF there would eventually be a representative from the local electrical board to investigate (how quickly depends on how much disturbances since this impacts the generation, the distribution and everyone at the PCC)
In practice you cannot have an ideal inductor and thus there would be real power being pulled to feed the resistive element of the inductor and the cableing while also drawing (potentially) excessive current and DPF