1. **Problem statement:** A man starts at 0 on the number line and makes 101 moves. On the $n$th move, he moves $n$ units either left or right. We want to find how close to 0 his final position can be after all 101 moves. 2. **Total distance moved:** The total distance moved is the sum of the first 101 natural numbers: $$\text{Total} = 1 + 2 + 3 + \cdots + 101 = \frac{101 \times 102}{2} = 5151.$$ 3. **Final position:** The final position is the difference between the sum of distances moved right and the sum of distances moved left. Let $S$ be the sum of distances moved right, then the sum moved left is $5151 - S$. The final position is: $$|S - (5151 - S)| = |2S - 5151|.$$ 4. **Goal:** We want to minimize $|2S - 5151|$ by choosing $S$ as a sum of some subset of $\\{1, 2, ..., 101\\}$. 5. **Key insight:** Since $5151$ is odd, $2S$ is always even, so $|2S - 5151|$ is always odd. The minimum positive odd number is 1, so the closest possible final position to 0 is 1. 6. **Conclusion:** The man cannot end exactly at 0 because the total sum is odd, but he can get as close as 1 unit from 0. **Final answer:** $$\boxed{1}$$