I am taking part in a robotics competition, where the challenge is to create a pair of robots which successfully navigate a series of obstacles. However, the rules state that of the two robots, only one must have a driving actuator. The other must somehow be moved by the other robot, WITHOUT PHYSICAL CONTACT.

I could think of either having sails on the non-driving robot, and moving it with fans on the driving one OR electromangnets on the driving one and permanent magnets with the opposite polarity on the non-driving one. However the problem with both is that efficiency falls off drastically with distance. Thus, I am looking for possible ways to overcome this problem.

Also, the driving robot has a cable power supply, while the non-driving one may only have batteries which can steer it.

I believe this question belongs to Physics as I want to know the actual forces possible, not their implementation in robots. Thanks :)
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This might be a bit helpful. 

The uncertainty in the momentum of an electron in an atom is defined as: $(\Delta P )^{2}= \langle P^{2}\rangle-\langle P\rangle^{2}$. An electron bounded by the nucleus, has an average momentum of zero, which means ($\Delta P)^2 = {\langle P^{2}\rangle}$, but again, Feynman's explain, as explained in the comment, was a really hand-wavy approach towards approximating the size of an atom.
In QM that electron would need more and more kinetic energy when it closes to nucleus, so at some point Coulombic force becomes too weak. The result is an equilibrium position with nucleus, i.e. hydrogen atom.

Heisenberg's uncertainty principle is useless when it comes to hydrogenic $1/r$ potential, but for harmonic oscillator $r^2$ potential we can use Heisenberg's uncertainty principle to easily calculate the ground state energy.

What you need is the following "uncertainty principle" that solves $1/r$ potential and the ground state energy for hydrogen atom, when $\psi$ is the wave for electron. You can prove it as an easy homework exercise:
$$\int\frac{|\psi(x,t)|^2}{|x|}dx \le \sqrt{\int |\nabla\psi(x,t)|^2dx}.$$