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 if all steps were in the same direction is the sum of the first 101 natural numbers: $$ S = 1 + 2 + 3 + \cdots + 101 = \frac{101 \times 102}{2} = 5151 $$ 3. **Final position:** The final position is the difference between the sum of steps taken to the right and the sum of steps taken to the left. Let the sum of steps to the right be $R$ and to the left be $L$. Then: $$ R + L = 5151 $$ $$ \text{Final position} = |R - L| $$ 4. **Goal:** Minimize $|R - L|$. Since $R + L = 5151$, we have: $$ |R - L| = |R - (5151 - R)| = |2R - 5151| $$ 5. **Finding minimum difference:** We want to find $R$ such that $|2R - 5151|$ is minimized. Since $5151$ is odd, the minimum difference is 1 because $2R$ is always even and cannot equal an odd number exactly. 6. **Conclusion:** The closest the man can get to 0 is 1 unit away. **Final answer:** 1