Find all pairs of odd integers $a$ and $b$ for which there exists a natural number$ c$ such that  the number $\frac{c^n+1}{2^na+b}$ is integer for all natural $n$.
1. **Initial Setup and Analysis:**
   We need to find pairs of odd integers \(a\) and \(b\) such that there exists a natural number \(c\) for which the expression \(\frac{c^n + 1}{2^n a + b}\) is an integer for all natural numbers \(n\).

2. **Choosing Specific Values for \(n\):**
   Let's start by choosing \(n = 2^t\) for all \(t\). This simplifies our expression to:
   \[
   \frac{c^{2^t} + 1}{2^{2^t} a + b}
   \]
   For this to be an integer, the denominator \(2^{2^t} a + b\) must divide the numerator \(c^{2^t} + 1\).

3. **Order Analysis:**
   By analyzing the order of the terms, we note that every odd prime factor of \(c^{2^t} + 1\) is congruent to \(1 \mod 2^{t+1}\). Since \(2^{2^t} a + b\) is odd, it must also be congruent to \(1 \mod 2^{t+1}\).

4. **Congruence Condition:**
   This gives us the congruence:
   \[
   2^{2^t} a + b \equiv 1 \mod 2^{t+1}
   \]
   Simplifying, we get:
   \[
   b - 1 \equiv 0 \mod 2^{t+1}
   \]
   This implies:
   \[
   2^{t+1} \mid b - 1
   \]
   Since this must hold for all \(t\), \(b - 1\) must be divisible by arbitrarily large powers of 2. Therefore, \(b = 1\).

5. **Simplifying with \(b = 1\):**
   Now, we need to check if there exists a natural number \(c\) such that:
   \[
   \frac{c^n + 1}{2^n a + 1}
   \]
   is an integer for all \(n\).

6. **Prime Factor Analysis:**
   For all primes \(p\), we have:
   \[
   2^p a + 1 \mid c^p + 1
   \]
   By Fermat's Little Theorem, every prime divisor of \(c^p + 1\) is either \(1 \mod p\) or divides \(c + 1\).

7. **Defining \(f(p)\):**
   Let \(f(p) = 2^p a + 1\). We denote \(s_p = \gcd(c + 1, f(p))\). By Fermat's Little Theorem:
   \[
   2a + 1 \equiv f(p) \equiv s_p \mod p
   \]
   Thus:
   \[
   p \mid 2a + 1 - s_p
   \]
   Since \(s_p \leq c + 1\), for sufficiently large \(p\), we must have \(s_p = 2a + 1\).

8. **Divisibility Condition:**
   For large primes \(p\), we have:
   \[
   2a + 1 \mid f(p)
   \]
   This implies:
   \[
   2a + 1 \mid a (2^{p-q} - 1)
   \]
   for primes \(p > q\). Since \(\gcd(a, 2a + 1) = 1\), it follows that:
   \[
   2a + 1 \mid 2^{p-q} - 1
   \]

9. **Order of 2 Modulo \(2a + 1\):**
   This implies \(2a + 1\) divides \(2^d - 1\) where \(d \mid p - q\) for all pairs of such primes. By Dirichlet's Theorem, for \(k \geq 2\), there are large primes \(p \equiv 1 \mod 2k\) and \(q \equiv -1 \mod 2k\), so \(2k \nmid |p - q|\). Thus, \(k = 1\) and:
   \[
   2a + 1 \mid 3
   \]
   This implies \(a = 1\).

10. **Verification:**
    For \((a, b) = (1, 1)\), we can take \(c = 2\) and verify:
    \[
    \frac{2^n + 1}{2^n + 1} = 1
    \]
    which is an integer for all \(n\).

The final answer is \( \boxed{ (a, b) = (1, 1) } \)