An eccentric mathematician has a ladder with $ n$ rungs that he always ascends and descends in the following way: When he ascends, each step he takes covers $ a$ rungs of the ladder, and when he descends, each step he takes covers $ b$ rungs of the ladder, where $ a$ and $ b$ are fixed positive integers. By a sequence of ascending and descending steps he can climb from ground level to the top rung of the ladder and come back down to ground level again. Find, with proof, the minimum value of $ n,$ expressed in terms of $ a$ and $ b.$
1. **Claim**: The minimum value of \( n \) is \( a + b - \gcd(a, b) \).

2. **Proof using Bézout's Lemma**:
   - Bézout's Lemma states that for any integers \( a \) and \( b \), there exist integers \( x \) and \( y \) such that \( ax - by = \gcd(a, b) \).
   - We can find non-negative integers \( x \) and \( y \) such that \( ax - by = \gcd(a, b) \).
   - Rearranging, we get \( b(y + 1) - a(x - 1) = a + b - \gcd(a, b) \).
   - This shows that it is possible to perform a sequence of moves that cover \( a + b - \gcd(a, b) \) rungs.

3. **Verification**:
   - Suppose \( n < a + b - 1 \).
   - Considering all the moves in total, we have \( ax = by \) for some \( x, y \in \mathbb{Z} \), implying that we move down at least \( a \) times and similarly move up at least \( b \) times.
   - Since \( \gcd(a, b) = 1 \), \( b \cdot k \) (for \( k \in \{0, 1, \dots, a-1\} \)) forms a complete residue system modulo \( a \), so we would necessarily have the height be \( b - 1 \pmod{a} \) at some point.
   - But since \( n < a + b - 1 \), we have no legal move at this point, which is a contradiction.

4. **Conclusion**:
   - Therefore, the minimum value of \( n \) is \( a + b - \gcd(a, b) \).

\(\blacksquare\)

The final answer is \( \boxed{ a + b - \gcd(a, b) } \)