a pigeon hole principle exercise

From Matousek's discrete math book.

Prove that any n+1 numbers from 1 to 2n contains two numbers a and b such that a divides b.

Some observations: first n+1 is necessary to guarantee the existence of a and b. If we only pick n numbers, then pick n+1 ... 2n will do. No two divides each other.

second is that you need pigeon hole principle, and you need to build n holes somehow. Several attempts do not seem to be promising. The idea suggested in the solution is to build buckets with respect to 2^k*(2*m+1), that is, use the odd factor to build buckets. Now you have 500 buckets, with odd factor 1, 3, 5, ..., 999. Numbers fall into the same bucket has the form
2*(2m+1), 2^2*(2m+1), ..., 2^k*(2m+1) and the min will divide the max.

Are there other proofs?