Regarding this question, I will try to use the analogy of the differential equations $$F = m\frac{dv}{dt}$$ and $$V = L\frac{dI}{dt}$$ and $$w = \int Fv dt$$ with $$w = \int VI dt$$ to figure out the concept of reactive power.
If I move a mass back and forth, so that its velocity changes like $$v = v_0sin(\omega t)$$ my hands are doing a force similar to what a spring would do. But in this case, there is no spring and no friction also.
Formally, if the product Fv is integrated along a cycle, the result is zero because part of the time they have the same sign, and part opposite sign. But if I am the source of that accelerated movement, I am doing work along the cycle, in the meaning that I am spending energy.
Now compare with a current, also changing in a sinusoidal way in an inductor and without resistance. There is a changing voltage associated with that current, and without any capacitor, this voltage must be supplied by the source. In this case it is also true that the work, as the integral of the electric power during the time of the cycle is zero.
But the source of the electrical energy (say a diesel generator) is burning fuel to keep the AC running in the circuit.
It is different in the case of a spring-mass or inductor-capacitor system. Without friction (resistance) the oscillations don't need a source (except for initiate the harmonic motion).
Is it correct this mechanical analogy to understand reactive power?