Let \(c\) be a positive real number. Alice wishes to pick an integer \(n\) and a sequence \(a_1\), \(a_2\), \(\ldots\) of distinct positive integers such that \(a_{i} \leq ci\) for all positive integers \(i\) and \[n, \qquad n + a_1, \qquad n + a_1 - a_2, \qquad n + a_1 - a_2 + a_3, \qquad \cdots\] is a sequence of distinct nonnegative numbers. Find all \(c\) such that Alice can fulfil her wish.
1. **Define the sequence and initial conditions:**
   Let \( b_k = n + a_1 - a_2 + a_3 - a_4 + \cdots + (-1)^{k-1} a_k \) with \( b_0 = n \). Define \( B_k = \{b_0, b_1, b_2, \ldots, b_k\} \) and \( A_k = \{a_1, a_2, \ldots, a_k\} \).

2. **State the lemma and its conditions:**
   We will show that we can extend the sequence \( (a_1, a_2, \ldots, a_k) \) by either 2, 4, or 6 elements while preserving the following conditions:
   - (a) \( k \) is even.
   - (b) The elements of \( A_k \) are all distinct, with \( a_{k-1} \) being the largest element.
   - (c) The elements of \( B_k \) are all distinct, with \( b_{k-1} \) being the largest element.
   - (d) Any pair of elements of \( B_k^* = B_k \cap (b_k, \infty) \) are at least 3 units apart.

3. **Case I: \( b_k + 1 \notin B_k^* \):**
   - Set \( k' = k + 2 \) with new elements \( (a_{k+1}, a_{k+2}) = (a_{k-1} + 2, a_{k-1} + 1) \).
   - Verify conditions:
     - \( b_{k+1} - b_{k-1} = a_{k+1} - a_k \geq 3 \) since \( a_{k+1} > a_{k+2} > a_{k-1} > a_k \).
     - \( b_{k+2} = b_k + a_{k+1} - a_{k+2} = b_k + 1 \), ensuring \( b_{k+1}, b_{k+2} \) are distinct from \( B_k \).
     - Conditions (a) and (b) are clearly preserved.

4. **Case II: \( b_k + 1 \in B_k^* \) but \( b_k + 4 \notin B_k^* \):**
   - Set \( k' = k + 4 \) with new elements \( (a_{k+1}, a_{k+2}, a_{k+3}, a_{k+4}) = (a_{k-1} + 3, a_{k-1} + 1, a_{k-1} + 4, a_{k-1} + 2) \).
   - Verify conditions (details left for the diligent solver).

5. **Case III: \( b_k + 1, b_k + 4 \in B_k^* \) but \( b_k + 7 \notin B_k^* \):**
   - Set \( k' = k + 6 \) with new elements \( (a_{k+1}, a_{k+2}, a_{k+3}, a_{k+4}, a_{k+5}, a_{k+6}) = (a_{k-1} + 3, a_{k-1} + 1, a_{k-1} + 5, a_{k-1} + 2, a_{k-1} + 6, a_{k-1} + 4) \).
   - Verify conditions (details left for the diligent solver).

6. **Case IV: \( b_k + 1, b_k + 4, b_k + 7 \in B_k^* \) but \( b_k + 10 \notin B_k^* \):**
   - Set \( k' = k + 6 \) with new elements \( (a_{k+1}, a_{k+2}, a_{k+3}, a_{k+4}, a_{k+5}, a_{k+6}) = (a_{k-1} + 4, a_{k-1} + 1, a_{k-1} + 5, a_{k-1} + 2, a_{k-1} + 6, a_{k-1} + 3) \).
   - Verify conditions (details left for the diligent solver).

7. **Construct the sequence for \( c > 1 \):**
   - Let \( N \) be an integer such that \( c > 1 + \frac{1}{N} \).
   - Pick an even, large \( M > N^2 + 6N + 1 \).
   - Set \( n = M \) and \((a_1, a_2, \ldots, a_{M-2}) = (1, 2, \ldots, N-2, N-1, N+1, N+2, \ldots, M-1)\) and \( (a_{M-1}, a_M) = (M + N, N) \).

8. **Verify the conditions for \( a_i \leq ci \):**
   - For \( i \leq N-1 \), \( \frac{a_i}{i} = 1 < c \).
   - For \( N \leq i \leq M-2 \), \( \frac{a_i}{i} = 1 + \frac{1}{i} \leq 1 + \frac{1}{N} < c \).
   - For \( i = M-1 \), \( \frac{a_i}{i} = 1 + \frac{N+1}{M-1} \leq 1 + \frac{1}{N} < c \).
   - For \( i = M \), \( \frac{a_i}{i} = \frac{N}{M} < 1 < c \).

9. **Extend the sequence using the lemma:**
   - Inductively extend the sequence such that \( a_i \leq i + N + 6 \) for all \( i > M \).
   - Thus, \( \frac{a_i}{i} \leq 1 + \frac{N+6}{i} \leq 1 + \frac{N+6}{M} < c \) for all \( i > M \).

10. **For \( c \leq 1 \):**
    - \( a_i \leq i \) are distinct positive integers, implying \( a_k = k \) for all \( k \in \mathbb{N} \).
    - Inductively, \( b_{2k} = n - k \) for all \( k \in \mathbb{N} \), which becomes negative for large \( k > n \), leading to a contradiction.

The final answer is \( \boxed{ c > 1 } \).