Show that in a subset of n + 1 distinct numbers from the set A =
{1, . . . , 2n}, there are always two distinct elements such that one divides the other.
When I don't have any idea what to do, I tend to "play" with the problem to get a feel for what it means.
Suppose n=2; they are claiming that any subset of 3 numbers from {1, 2, 3, 4} includes at least one pair, one of which is a multiple of the other. The possible subsets of 3 are {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}. The first three obviously work, and {2, 3, 4} has the subset {2, 4}.
You might try n=3, and see if there's a pattern in what is divisible by what. (For example, I wondered at first if there would always be a pair with the ratio 2:1, but {1, 3, 4} shows I was wrong.)
You might also think about methods you've learned. This sounds as if it could be a pigeonhole problem; is that something you're studying?
Now please show some work, so we can start actually helping!