1. **Problem Statement:** We need to show that if we choose 101 numbers from the list 1 to 200, there must be at least two numbers among the chosen that have a highest common factor (HCF) of 1. 2. **Key Idea:** The HCF of two numbers is the greatest number that divides both. If two numbers are consecutive integers, their HCF is always 1. 3. **Important Rule:** For any two consecutive integers $n$ and $n+1$, $$\text{HCF}(n, n+1) = 1.$$ This is because consecutive numbers differ by 1, and no number greater than 1 can divide both. 4. **Applying the Pigeonhole Principle:** The list from 1 to 200 has 199 pairs of consecutive integers: (1,2), (2,3), ..., (199,200). 5. If we select 101 numbers from 1 to 200, by the pigeonhole principle, at least one pair of these chosen numbers must be consecutive. This is because the maximum number of integers you can choose without picking any two consecutive numbers is 100 (e.g., choosing all odd numbers). 6. Since at least one pair of chosen numbers is consecutive, their HCF is 1. 7. **Conclusion:** Therefore, among any 101 numbers chosen from 1 to 200, there exist at least two numbers whose HCF is 1. **Final answer:** There must be at least two numbers with HCF equal to 1 among any 101 numbers chosen from 1 to 200.