Given a (fixed) positive integer $N$, solve the functional equation
\[f \colon \mathbb{Z} \to \mathbb{R}, \ f(2k) = 2f(k) \textrm{ and } f(N-k) =  f(k), \ \textrm{for all } k \in \mathbb{Z}.\]

[i](Dan Schwarz)[/i]
To solve the functional equation 
\[ f \colon \mathbb{Z} \to \mathbb{R}, \ f(2k) = 2f(k) \textrm{ and } f(N-k) = f(k), \ \textrm{for all } k \in \mathbb{Z}, \]
we will proceed step-by-step.

1. **Using the first condition \( f(2k) = 2f(k) \):**
   \[ f(2k) = 2f(k). \]

2. **Using the second condition \( f(N-k) = f(k) \):**
   \[ f(N-k) = f(k). \]

3. **Combining both conditions:**
   \[ f(2k) = 2f(k) = 2f(N - k). \]

4. **Using the periodicity property:**
   \[ f(2k) = f(2N - 2k). \]
   \[ f(2k) = f(N - (2k - N)) = f(2k - N). \]

5. **By induction, we can show that \( f(2k - rN) = f(2k) \) for all integers \( r \):**
   \[ f(2k - rN) = f(2k). \]

6. **Choosing \( r = 2s \):**
   \[ f(2k - 2sN) = 2f(k - sN). \]
   \[ f(2k - 2sN) = f(2k) = 2f(k). \]

7. **Thus, \( f \) is periodic with period \( N \):**
   \[ f(k + N) = f(k). \]

8. **Now, suppose \( N \) is even:**
   \[ N = 2m. \]
   \[ f(2k) = f(2k + N) = f(2(k + m)) = 2f(k + m). \]
   \[ 2f(k) = 2f(k + m). \]
   \[ f(k) = f(k + m). \]

9. **Thus, \( f \) is periodic with period \( N/2 \):**
   \[ f(k + N/2) = f(k). \]

10. **By induction, \( f \) is periodic with an odd period, say \( b \):**
    \[ f(a2^\varphi(b)) = f(a) \]
    by Euler's theorem and
    \[ f(a2^\varphi(b)) = 2^\varphi(b)f(a) \]
    by the given condition.

11. **Thus, \( f(a) = 0 \) for all integers \( a \):**
    \[ f(a) = 0. \]

The final answer is \( \boxed{ f(a) = 0 } \) for all integers \( a \).