Let $k$ be a positive integer.  Determine the least positive integer $n$, with $n\geq k+1$, for which the game below can be played indefinitely:

Consider $n$ boxes, labelled $b_1,b_2,...,b_n$. For each index $i$, box $b_i$ contains exactly $i$ coins. At each step, the following three substeps are performed in order:
[b](1)[/b] Choose $k+1$ boxes;
[b](2)[/b] Of these $k+1$ boxes, choose $k$ and remove at least half of the coins from each, and add  to the remaining box, if labelled $b_i$, a number of $i$ coins.
[b](3)[/b] If one of the boxes is left empty, the game ends; otherwise, go to the next step.

[i]Proposed by Demetres Christofides, Cyprus[/i]
1. **Initial Setup and Definitions:**
   - Let \( k \) be a positive integer.
   - We need to determine the least positive integer \( n \) such that \( n \geq k+1 \) and the game can be played indefinitely.
   - The game involves \( n \) boxes labeled \( b_1, b_2, \ldots, b_n \), where box \( b_i \) contains exactly \( i \) coins initially.
   - At each step, the following substeps are performed:
     1. Choose \( k+1 \) boxes.
     2. From these \( k+1 \) boxes, choose \( k \) and remove at least half of the coins from each, and add to the remaining box (if labeled \( b_i \)) a number of \( i \) coins.
     3. If one of the boxes is left empty, the game ends; otherwise, go to the next step.

2. **Finding the Minimum \( n \):**
   - We claim that the least positive integer \( n \) for which the game can be played indefinitely is \( n = 2^k + k - 1 \).

3. **Proof that \( n = 2^k + k - 1 \) Works:**
   - Let \( a_0 \leq a_1 \leq \cdots \leq a_k \) be the numbers of coins in the \( k+1 \) largest-indexed boxes, in sorted order.
   - Define a position as *full* if \( a_i \geq 2^i \) for each \( 0 \leq i \leq k \).
   - Initially, the position is full because the number of coins in each box \( b_i \) is \( i \), and the indices of the \( k+1 \) largest boxes are at least \( 2^k - 1 \).
   - If we add to the box with \( a_0 \) coins and subtract from the other \( k \) boxes, fullness is preserved.

4. **Proof that \( n < 2^k + k - 1 \) Does Not Work:**
   - Define \( f(n) \) as the greatest nonnegative integer \( r \) such that \( 2^r \leq n \).
   - If box \( i \) has \( c_i \) coins, let the *score* of the position be \( \sum f(c_i) \) over all boxes \( i \).

5. **Claim:**
   - \( f(x + y - 1) \leq f(x) + f(y) \).

6. **Proof of the Claim:**
   - If \( x \) or \( y \) is \( 1 \), the claim is clearly true.
   - Otherwise, let \( f(x) = r \) and \( f(y) = s \), with \( r \geq s \geq 1 \).
   - Then:
     \[
     x + y - 1 < 2^{r+1} + 2^{s+1} - 1 < 2^{r+1} + 2^{r+1} = 2^{r+2} \leq 2^{r+s+1}
     \]
   - This implies \( f(x + y - 1) \leq r + s \).

7. **Score Analysis:**
   - If we remove at least half the coins from box \( i \), this subtracts at least \( 1 \) from \( f(c_i) \).
   - If we add \( i \) coins to box \( i \), by the above claim, this adds at most \( f(i + 1) \leq k \) coins to box \( i \), and equality holds for at most \( k \) values of \( i \).
   - Therefore, the score can never increase in a step, and there are at most \( k \) boxes (chosen as the box being added to) for which the score remains the same.

8. **Conclusion:**
   - The score is always a nonnegative integer, so after some point, we must only add to one of those \( k \) boxes.
   - This means we must subtract from one of the remaining \( n - k \) boxes, causing the number of coins in those \( n - k \) boxes to strictly decrease every step, leading to a contradiction.

The final answer is \( \boxed{ n = 2^k + k - 1 } \).