$n \ge 4$ real numbers are arranged in a circle. It turned out that for any four consecutive numbers $a, b, c, d$, that lie on the circle in this order, holds $a+d = b+c$. For which $n$ does it follow that all numbers on the circle are equal?

[i]Proposed by Oleksiy Masalitin[/i]
1. **Restate the problem condition**: We are given \( n \ge 4 \) real numbers arranged in a circle such that for any four consecutive numbers \( a, b, c, d \) in this order, the condition \( a + d = b + c \) holds. We need to determine for which \( n \) it follows that all numbers on the circle are equal.

2. **Analyze the condition for even \( n \)**: 
   - Consider \( n \) to be even. We can construct a counterexample with the sequence \( 1, 2, 1, 2, \ldots, 1, 2 \) arranged in a circle. 
   - For any four consecutive numbers \( a, b, c, d \) in this sequence, we have \( a = 1 \), \( b = 2 \), \( c = 1 \), and \( d = 2 \). 
   - Checking the condition: \( a + d = 1 + 2 = 3 \) and \( b + c = 2 + 1 = 3 \), which satisfies \( a + d = b + c \).
   - However, not all numbers are equal. Thus, for even \( n \), it is possible to have numbers that are not all equal.

3. **Analyze the condition for odd \( n \)**:
   - Let \( a_1, a_2, \ldots, a_n \) be the \( n \) numbers arranged in a circle.
   - The given condition \( a + d = b + c \) can be rewritten in terms of indices as \( a_{i} + a_{i+3} = a_{i+1} + a_{i+2} \) for all \( i \), where indices are taken modulo \( n \).
   - Rearrange this to \( a_{i+3} - a_{i+1} = a_{i+2} - a_{i} \). Let \( k \) be the common difference, so \( a_{i+2} - a_{i} = k \) for all \( i \).

4. **Establish the common difference**:
   - From the above, we have \( a_3 - a_1 = k \), \( a_4 - a_2 = k \), \( a_5 - a_3 = k \), and so on.
   - This implies \( a_{i+2} - a_i = k \) for all \( i \). Since \( n \) is odd, we can cycle through all indices and eventually return to the starting point.
   - Summing these differences over one complete cycle: 
     \[
     (a_3 - a_1) + (a_4 - a_2) + \ldots + (a_1 - a_{n-1}) + (a_2 - a_n) = kn
     \]
   - Since the sum of differences over a complete cycle must be zero (as we return to the starting point), we have \( kn = 0 \). Given \( n \neq 0 \), it follows that \( k = 0 \).

5. **Conclude that all numbers are equal**:
   - Since \( k = 0 \), we have \( a_{i+2} - a_i = 0 \) for all \( i \), implying \( a_{i+2} = a_i \).
   - This means \( a_1 = a_3 = a_5 = \ldots = a_n \) and \( a_2 = a_4 = a_6 = \ldots = a_{n-1} \).
   - Since \( n \) is odd, the sequences \( a_1, a_3, \ldots, a_n \) and \( a_2, a_4, \ldots, a_{n-1} \) overlap, implying all \( a_i \) are equal.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) must be odd.