energy of eddy current with increasing frequency As far as I know, eddy current losses increase with the increase of change of magnetic field.
That means that they increase with a square of a peak of magnetic field and with a square of frequency.
Now what I don't understand is from where does the energy come from if I just increase the frequency of the power delivery system. The energy input is the same at both frequencies. But the eddy current losses actualy increase with a square of frequency.
That means that at some point you could excede the 100 % efficiency. Why this doesn't happen?
 A: In detail eddy currents are very complicated and a 100% answer might not be possible. First of all, it should be clear that the efficiency will not be greater than 100%. 
The Eddy current is taking energy from your system. This is due to the fact that your osculating field is actually radiating. (have a look at Pointing Vectors) This energy loss is small compared to, e.g., ohmic losses, so you do not care very much. If you could increase frequency to any arbitrary value, your radiated energy will increase, eventually limiting your energy input, i.e. there will probably be a max frequency.
A second point is to consider that the current distribution changes with frequency.The article on Eddy Currents mentions for the formula, you are probably talking about, neglecting the Skin Effect.
Finally, you have to consider that large eddy currents will produce very large fields themselves, which then will couple in your field generating coil and power system, making things even more complicated.
 Long story short: For ever increasing frequencies your $E \propto f^2$ formula becomes invalid.
Edit
Due to the ongoing discussion in the comments, let me try to give another hint.You can look at your system: 
Power source $\rightarrow$ Coil $\rightarrow$ Field $\rightarrow$ Metal object $\rightarrow$ Eddy current $\rightarrow$ Loss
as a transformer with load. The metal object is the second coil, the Eddy current the current flowing if a load is provided and the resistance acting on the Eddy current the load. 
