Hi, I very appreciate your answer.
What do you consider as the USB part?
The circuitry inside the IC that deals with the USB signals, not the power supplies (ie. internal regulator). The manufacturers could have completely removed the regulator from the IC and it would still have all the USB functions, except you would have to provide a 3.3V regulator separately. See Figure 1 in your datasheet. The USB part is every block that is not the internal regulator (well technically the UART is a separate part too because it serves a different function but it is also powered from the internal regulator).
The USB supplies voltage for the regulator, meaning the USB operates the regulator, so how come the regulator runs the USB part?
Isnt it true that the USB is the power supply of the whole chip?
And therefore the current that the regulator sources is coming from the USB?
You are getting confused about two separate "parts" being housed in a single IC pacakge. The difference is as meaningless as to whether or not your PC power supply was mounted inside your tower, or sitting outside beside it. Like I said imagine your one IC as TWO separate ICs (a USB IC and a regulator IC). The one with all the USB functionality and a linear regulator. The manufacturers just threw in the regulator with the USB part so you don't have to juggle two ICs in order to make things more convenient for you. The PC's USB powers the regulator which in turn powers the USB part of the chip. So yes, the PC's USB powers the "whole chip" BUT the whole chip consists of the regulator part and USB part. The PC's USB directly powers the internal regulator, and the internal regulator directly powers the USB part, therefore the PC's USB only indirectly powers the USB part. Understand? Remember think of them as two separate parts.
Now, the reason the regulator part of the IC runs the USB part of the IC Is because the PC USB outputs 5V, but the USB part of the IC needs 3.3V. Look at Figure 1 in your data sheet, you can see that the regulator block accepts 5V and turns it into 3.3V which powers everything else EXCEPT the USB transceiver which has it's own separate connection to 5V. So internally, the IC processes the information using 3.3V (because it lets them uses lower power or faster transistors, and lets the IC communicate over the UART with 3.3V microcontrollers), but when the IC needs to send data to the PC it converts the signal back to 5V (because USB signals are 5V). So, in fact a tiny section of the USB part is powered directly from the PC's USB, but most if it is powered indirectly from the PC's USB (powered directly from the internal regulator and indirectly through the PC's USB).
I'm just trying to get you to visualize the functional components of the chip. Visualizing the chip as a whole is meaningless because you can stuff as many components as you want into a single IC package.
Assuming that it is a linear regulator, then IIN =~ IOUT in the regulator, so how come the regulator sources 26mA but sinks only µA?
26mA + XXXuA enter the IC, but the 26mA is not really consumed (sink may the wrong word to use here, sorry about that). Some of the extra voltage is removed to bring it down to 3.3V, but it exits the IC at 3.3V (not at 0V) with enough energy left to power lots of stuff (the energy was not consumed by the internal regulator).