The nuclear Overhauser effect
Consider the simplest possible NOE experiment, with two dipole-coupled protons, the 
target I and nearby proton S. 

Before the experiment there are four populated spin states: $\alpha_{I}\alpha_{S}$, $\beta_{I}\alpha_{S}$, $\alpha_{I}\beta_{S}$ and $\beta_{I}\beta_{S}$, where $\alpha$ indicates $m_{i} = \frac{1}{2}$ and $\beta$ indicates $m_{i} = -\frac{1}{2}$. The relative populations of the states at thermal equilibrium (from Boltzmann distribution) are:


$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}$,  $\alpha_{I}\beta_{S}: 1$
$\beta_{I}\beta_{S}: 1 - 2\Delta $


S is irradiated so that the population difference between the $m_{S} = \pm \frac{1}{2}$ states goes to zero. This results immediately in the following populations: 


$\alpha_{I}\alpha_{S}, \beta_{I}\alpha_{S}: 1 + \Delta $
$\beta_{I}\beta_{S}, \alpha_{I}\beta_{S}: 1 - \Delta $


What happens next typically for small molecules is fast "flip-flip" cross-relaxation, restoring the $\alpha_{I}\alpha_{S}, \beta_{I}\beta_{S} $ states to their equilibrium populations. So we have the populations:


$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}: 1 + \Delta$
$\alpha_{I}\beta_{S}: 1 - \Delta $
$\beta_{I}\beta_{S}: 1 - 2\Delta $


The population difference between the pairs $\alpha_{I}\beta_{S}, \alpha_{I}\alpha_{S}$ and  $\beta_{I}\beta_{S}, \beta_{I}\alpha_{S}$ has now increased by 50%, from $2\Delta$ to $3\Delta$. This corresponds to a 50% increase in intensity of the peak. Transition between these pairs obviously corresponds to spin-lattice relaxation.