Find the least $n$ such that any subset of ${1,2,\dots,100}$ with $n$ elements has 2 elements with a diﬀerence of 9.
1. We need to find the smallest \( n \) such that any subset of \(\{1, 2, \ldots, 100\}\) with \( n \) elements has at least two elements with a difference of 9.
2. Consider the set \(\{1, 3, 5, 7, \ldots, 99\}\). This set contains all odd numbers from 1 to 99. There are 50 elements in this set.
3. Notice that no two elements in this set differ by 9. For example, \(1\) and \(3\) differ by \(2\), \(3\) and \(5\) differ by \(2\), and so on. The difference between any two elements in this set is always an even number, and 9 is an odd number.
4. Therefore, a subset of \(\{1, 2, \ldots, 100\}\) with 50 elements can be constructed such that no two elements have a difference of 9.
5. Now, consider any subset of \(\{1, 2, \ldots, 100\}\) with 51 elements. By the pigeonhole principle, since there are only 50 possible remainders when dividing by 9 (i.e., \(0, 1, 2, \ldots, 8\)), at least two of these 51 elements must have the same remainder when divided by 9.
6. If two elements have the same remainder when divided by 9, their difference is a multiple of 9. Since we are working within the set \(\{1, 2, \ldots, 100\}\), the only multiple of 9 that is less than 100 is 9 itself.
7. Therefore, any subset of \(\{1, 2, \ldots, 100\}\) with 51 elements must contain at least two elements whose difference is 9.

Conclusion:
The least \( n \) such that any subset of \(\{1, 2, \ldots, 100\}\) with \( n \) elements has two elements with a difference of 9 is \(\boxed{51}\).