Let $N$ be a positive integer and $A = a_1, a_2, ... , a_N$ be a sequence of real numbers.
Define the sequence $f(A)$ to be
$$f(A) = \left( \frac{a_1 + a_2}{2},\frac{a_2 + a_3}{2}, ...,\frac{a_{N-1} + a_N}{2},\frac{a_N + a_1}{2}\right)$$
and for $k$ a positive integer define $f^k (A)$ to be$ f$ applied to $A$ consecutively $k$ times (i.e. $f(f(... f(A)))$)
Find all sequences $A = (a_1, a_2,..., a_N)$ of integers such that $f^k (A)$ contains only integers for all $k$.
1. **Claim: There exists some integer \( L \) such that \( f^L(A) = \{c, c, c, \dots , c\} \) for some constant integer \( c \).**

   **Proof:**
   Let \( M_k \) and \( m_k \) denote the maximum and minimum elements respectively of the sequence \( f^k(A) \). Let \( a^k_i \) denote the \( i \)-th element of the sequence \( f^k(A) \), where indices are taken \(\pmod{N}\). Due to the recursive relation of the sequences, we have:
   \[
   M_{k+1} = \dfrac{a^k_i + a^k_{i+1}}{2} \le \dfrac{M_k + M_k}{2} = M_k
   \]
   Similarly, we have that:
   \[
   m_{k+1} \ge m_k
   \]
   If \( M_{k+1} = M_k \) holds for infinitely many \( k \), that would imply that \( M_k = m_k \) so the sequence \( f^k(A) \) is constant, as desired. Thus, if we have that \( M_{k+1} = M_k \) holds for only finitely many \( k \), the sequence \( \{M_k\} \) is decreasing and bounded from below, so it must eventually be constant. A similar argument applies for \( \{m_k\} \).

2. **Case 1: \( N \) is odd**

   If \( N \) is odd, we have that \( f^{L-1}(A) = \{e, f, e, \dots , e\} \) for integers \( e, f \) such that \( e+f = 2c \). But then \( e+e = 2c \) so \( e = c \) and \( f = c \). Thus, working backwards, we have that:
   \[
   f(A) = f^{L-1}(A) = \{c, c, \dots, c\}
   \]

3. **Case 2: \( N \) is even**

   If \( N \) is even, we have that \( f^{L-1}(A) = \{e, f, e, \dots , f\} \) for integers \( e, f \) such that \( e+f = 2c \). Without loss of generality, let \( f \ge e \). Then we have:
   \[
   a^{L-2}_{2k} + a^{L-2}_{2k+1} = 2f \ge 2e = a^{L-2}_{2k+1} + a^{L-2}_{2k+2} \Rightarrow a^{L-2}_{2k} \ge a^{L-2}_{2k+2} \Rightarrow a^{L-2}_2 \ge a^{L-2}_N
   \]
   But we have:
   \[
   a^{L-2}_1 + a^{L-2}_2 = 2e \le 2f = a^{L-2}_1 + a^{L-2}_N \Rightarrow a^{L-2}_2 \le a^{L-2}_N
   \]
   So either \( a^{L-2}_2 = a^{L-2}_N \) so \( e = f = c \) or the sequence \( f^{L-2}(A) \) does not exist. Then we must have that:
   \[
   f(A) = \{e, f, e, \dots, f\}
   \]
   where \( e+f = 2c \) for some integer \( c \).

The final answer is \( \boxed{f(A) = \{e, f, e, \dots, f\}} \) where \( e+f = 2c \) for some integer \( c \).