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 (and here, for English web page context) 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 :-)
