Let $X_1, X_2, ... , X_n$ be a sample drawn without replacement from a finite population. $X_1$ may be the random variable - weight of the first person; $X_2$ may be the random variable - weight of the second person, etc. Then the random variables $X_1, X_2, ... , X_n$ are not mutually independent, but they are still said to be identically distributed. I can't make sense of why we'd think of $X_1, X_2, ... , X_n$ as identically distributed. It seems to defy intuition. Intuitively, $X_1, X_2, ... , X_n$ don't even necessarily have the same sample space, but in some sense this intuition must be incorrect. In what sense are $X_1, X_2, ... , X_n$ identically distributed (have the same marginal distribution)?
Let's look at the Wikipedia definition of marginal distribution.
In probability theory and statistics, the marginal distribution of a subset of a collection of random variables is the probability distribution of the variables contained in the subset. It gives the probabilities of various values of the variables in the subset without reference to the values of the other variables. This contrasts with a conditional distribution, which gives the probabilities contingent upon the values of the other variables.
The key sentence here is "It gives the probabilities of various values of the variables in the subset without reference to the values of the other variables." BUT, $X_2$ is defined with reference to the first draw. $X_2$ is the weight of the second person after we've drawn the first person and kept this first person out, making it unavailable to be drawn as the second person. It appears that the definition of marginal distribution assumes independent draws! And if that is true then it assumes sampling with replacement and is altogether useless (it seems) in this case as it appears to be perfectly contrary (assumes sampling with replacement) to the setup of this question (sampling without replacement)?