As pointed out by @loopy wait, this answer is wrong. Just leaving it up here for inspiration.
The answer is
A plane can budget its fuel as follows: 18L for flying away from port, 36L donated to the next furthest plane, 18L for flying towards port. This creates a "corridor"... imagine a plane at 0deg, 2 at 36deg, 2 at 72 deg, etc. From here, 2 planes fly towards each other to 18deg, 2 planes to 54 deg, two planes to 90 etc. Then they fly back to previous positions. The fuel is given by the previous plane in the loop. It should look like the picture below:
As you could calculate, the planes are only using 72L of their tank. This allows them to carry 18L, which it can use to escort another plane and donate 18L when turning back 18deg later. Hence you can fly the main plane 18*numRefuelers degrees.
It's basically the same thing as before, but without having to escort the main plane, the planes can seperate 22.5 degrees instead of 18deg, allowing
Starting the main plane:
We are going to use 8 refuelers with the corridor property, but the 8th will leave the main plane at 135deg, so it can fly solo till 225.
Retriving the refuelers:
When the main , we have a refueler at 9,27,45,63,81,99,117,135. Since there is no longer an escort, the fuel can be proportioned as 22.5L West, 22.5L East, 45L donated, forming a sustainance loop, freeing up the closest 2 planes to land ASAP (because 22.5*6=135 in a similar loop to the corridor) plus one plane clear to land every 22.5 minutes.
Recieving the plane:
27 minutes used, 63 till the solo is complete. No plane can fly to -135deg from the airport in 63 minutes. Therefore, when the first western refueler lands (81 minutes left), 3 refuelers must be in sustainance in -22.5, -45 and -67.5. With each western refueler landing and takes off to the east, the 3 planes continue the sustainance paths, but the paths themselves move further to the east by 22.5 degrees. Once the range of furthest eastern refueler reaches 135, the planes on the East (of which there should be 6 now) should do "corridor". However, note that because there are 7 planes (the main plane has been recieved) and 6 half-full planes able to provide fuel, the system has 240 L of fuel, holding the sustainance paths for 240/7~34 minutes before the planes run out of extra fuel and the last plane must be ditched.
The final stretch:
However, there are still planes on the west that are landing. If we can get 8 planes on the East within the 34 minutes we could land everyone safely. The 6th plane landed 13.5 minutes before the main solo flight ended, leaving 9 minutes till a 7th and 31.5 minutes till an 8th plane can be put on the East. This is less than 34 so we are saved! Now, we have the same corridor system as was used to start the professor, except in reverse. This allows the professor to land, as well as all their associates.
I apologise this is really confusing to read. I will try to make an animation soon, but I am very busy with school atm.