According to Regression Analysis by Example, the residual is the difference between response and predicted value, then it is said that every residual has different variance, so we need to consider standardized residuals.
But the variance is for a group of values, how could a single value have variance?
I would say that an individual number (such as a residual), which resulted from a random draw from a probability distribution, is a realized value, not a random variable. Likewise, I would say that the set of $N$ residuals, calculated from your data and your model fit using $\bf{e}=\bf{y}-\bf{\hat{y}}$, is a set of realized values. This set of numbers may be loosely conceptualized as independent draws from an underlying distribution $\epsilon$ ~ $\mathcal{N}(\mu,\sigma^2)$. (Unfortunately however, there are several additional complexities here. For example, you do not actually have $N$ independent pieces of information, because the residuals, $\bf{e}$, must satisfy two conditions: $\sum e_i=0$, and $\sum x_ie_i=0$.)
Now, given some set of numbers, be they residuals or whatever, it is certainly true that they have a variance, $\sum(e_i-\bar{e})^2/N$, but this is uninteresting. What we care about is being able to say something about the data generating process (for instance, to estimate the variance of the population distribution). Using the preceding formula, we could give an approximation by replacing the $N$ with the residual degrees of freedom, but this may not be a good approximation. This is a topic that can get very complicated very fast, but a couple of possible reasons could be heteroscedasticity (i.e., that the variance of the population differs at different levels of $x$), and the presence of outliers (i.e., that a given residual is drawn from a different population entirely). Almost certainly, in practice, you will not be able to estimate the variance of the population from which an outlier was drawn, but nonetheless, in theory, it does have a variance. I suspect something along these lines is what the authors had in mind, however, I should note that I have not read that book.
Update: Upon rereading the question, I suspect the quote may be referring to the way the $x$-value of a point influences the fitted regression line, and thereby the value of the residual associated with that point. The key idea to grasp here is leverage. I discuss these topics in my answer here: Interpreting plot.lm().
