Clearly if we can find largest, we can also find smallest so the solution is worded for largest. If we can do it for $4$ stones, we can do it for any $n\ge4$ stones; just keep on taking largest out of any $4$ until you're left with a winner. The case with $4$ stones should be easy casework.

@above: If you have 4 stones, it's not always possible to find the heaviest! Or, the lightest. But, as the statement claims, if you couldn't determine the former one, you will know which one is the lightest one, and vise versa. So, for 10 stones you can determine the two heviest ones, for example by doing all possible weighs. Removing them, you next find the two lightest, among the remaining stones. Then, by comparing the four chosen, you'll find either heaviest or lightest. It's an easy to see how.