Here is a quick answer...
Standard illustrative example
Let $y = (y_1, \dotsc, y_n)$ be a sample from a normal distribution $\mathrm{N}(\mu, \sigma^2$). Both $\mu$ and $\sigma^2$ are unknown. The maximum likelihood estimator of $\sigma^2$, obtained by taking the derivative of the log-likelihood with respect to $\sigma^2$ and equating to zero, is
$$
\hat{\sigma}^2_{\textrm{ML}} = \frac{1}{n} \sum_{i=1}^n (y_i -\bar{y})^2
$$
where $\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i$ is the maximum likelihood estimator of $\mu$. We can show that
$$
\mathrm{E}(\hat{\sigma}^2_{\textrm{ML}}) = \frac{n-1}{n} \sigma^2.
$$
[Start by rewriting $\hat{\sigma}^2_{\textrm{ML}}$ as $\frac{1}{n} \sum_{i=1}^n \left((y_i - \mu) + (\mu - \bar{y})\right)^2$]. Thus, $\hat{\sigma}^2_{\textrm{ML}}$ is biased. Note that if we had known $\mu$, then the MLE for $\sigma^2$ would have been unbiased. Hence, the problem with $\hat{\sigma}^2_{\textrm{ML}}$ appears to be linked with the fact that we have substituted $\bar{x}$ for the unknown mean in the estimation. The intuitive idea of REML estimation is to end up with a likelihood that contains all the information on $\sigma^2$ but no longer contains the information on $\mu$.
More technically, the REML likelihood is a likelihood of linear combinations of the original data: instead of the likelihood of $y$, we consider the likelihood of $Ky$, where the matrix $K$ is such that
$\mathrm{E}[Ky] = 0$.
REML estimation is often used in the more complicated context of mixed models. Every book on mixed models have a section explaining REML estimation in more details.
Edit
@Joe King: Here is one of my favorite books on mixed models that is fully available online. Section 2.4.2 deals with estimating variance components. Enjoy your reading :-)
