Power supply - Boost voltage over distances
It is difficult to build converters that have significantly greater than 90% conversion efficiency, especially when you have large conversion ratios (1:20 and 50:1, in this case). Therefore, even if you have no losses in the wire, you've already got an overall efficiency that's no greater than about 81%.

If you just transmit the 24 VDC across the 300 meters, and use a point-of-load converter to produce the 9V, the current in the long wire would be about 1.6 A. A 2-wire cable would have a total conductor length of 600 meters, and if we want to match the performance of the boost converter, it must drop no more than 2.4 V, which means that the total resistance can be no more than 1.5 &Omega;, or 2.5 m&Omega;/m. You would need to use AWG 8 wire (2 m&Omega;/m) in order to meet this requirement.

If you boost the voltage to 480 VDC, then the current in the wire is just 80 mA. AWG 18 wire (20.4 m&Omega;/m) would give you a voltage drop of about 1V (a loss of just 0.2%), so perhaps this voltage is overkill.

A voltage of 120 VDC would give you a current of 320 mA and a voltage drop of 4 V (3.3%) over that same AWG 18 wire, so perhaps this is a more reasonable value to use, giving you an overall efficiency on the order of 78%.