...and it's likely to spark some debate. Let me introduce the idea by way of an analogy. Let's say you're tired of the city and want to visit the country. Your friend agrees to lend you her car for the weekend so you can get away and clear your head. When you pick up the car from her, it has 3/4 of a tank of gas. The first thing you do is stop at the gas station to fill the tank and buy some snacks for the trip. The trip is great, and you return on Sunday evening completely refreshed. When you return the car, it has a little more than 3/4 of a tank of gas - say 7/16. Your friend notices you haven't used any of her gasoline, and says you can borrow the car again whenever you want.
After a few trips like this you decide it makes sense for you to buy your own car. Unfortunately, you are taken in by the slick salesman and end up with a car which has trouble starting. It turns out that the real issue is not the battery but the alternator - for some unknown reason, the alternator will keep the car running once it has started, but never re-charges the battery, not even a little. Luckily, your friend has a 67F capacitor she can lend you to start your car. When you pick up the capacitor it is charged to 12V, so contains 4824J of energy. Just like you did with the gasoline, you first "fill up" the capacitor to 14V (you happen to have a little device which does this) - now the capacitor contains 6566J. You connect the capacitor in place of your car battery, and start the car. Starting the car takes 1500J, so afterwards, the capacitor contains 5066J of energy, and is charged to 12.297V. Just as you did with her car, you return the capacitor to her with a little more energy than when you borrowed it. Again, she notices that you haven't used any of the energy in her capacitor, and offers to lend it to you whenever you need.
Unfortunately, the little device you have won't charge the capacitor from 0 to 14V, only from 12 to 14. So, if your friend lends you a fully discharged capacitor, you can't start your car. If, on the other hand, she lends you the same capacitor you began with, you always leave it with a little more energy than the last time you used it, and you can continue starting your car indefinitely using just the energy from your little device.
I don't know if I can extract 6566J from a coin cell fast enough to charge a supercapacitor, but I may be able to extract 1500J. In this case, I propose to begin with a capacitor charged to 12V, charge the capacitor to 14V, then start the car with it. By monitoring the voltage and current during starting with an oscilloscope, I can verify that the starting took less energy than was deposited in the capacitor by the coin cell. The numbers used here are just examples: the actual values will probably differ somewhat.
Ideally, I'd modify the car so that the alternator would not charge the capacitor once the engine is started (imagine a beefy ideal diode made with MOSFETs). The problem is that cars depend on the battery to filter the alternator voltage - with older cars, you could disconnect the battery once the car was running, but with modern computerized engines, this is asking for a whole lot of trouble. I am willing to believe that a 67F capacitor can stand in for the battery as a power supply filter, but don't think it is wise to have the capacitor isolated from the alternator once the engine is running. So, the next best thing is to analyze the voltage and current waveforms during starting to verify that none of the original charge of the capacitor was used in starting the engine.
Will the judges accept this argument? Will they disqualify my entry? What do you think?
Then again, maybe I can still find a way to get 6566J from a coin cell :-)