A function $f$ from the positive integers to the positive integers is called [i]Canadian[/i] if it satisfies $$\gcd\left(f(f(x)), f(x+y)\right)=\gcd(x, y)$$ for all pairs of positive integers $x$ and $y$.

Find all positive integers $m$ such that $f(m)=m$ for all Canadian functions $f$.
1. **Restate the problem and define the function:**
   A function \( f \) from the positive integers to the positive integers is called *Canadian* if it satisfies 
   \[
   \gcd\left(f(f(x)), f(x+y)\right) = \gcd(x, y)
   \]
   for all pairs of positive integers \( x \) and \( y \). We need to find all positive integers \( m \) such that \( f(m) = m \) for all Canadian functions \( f \).

2. **Transform the given equation:**
   By substituting \( y \rightarrow y - x \) in the given equation, we get:
   \[
   \gcd(f(f(x)), f(y)) = \gcd(x, y - x)
   \]
   for \( y > x \). Let this equation be denoted as \( P(x, y) \).

3. **Define magical numbers:**
   Call each number \( m \) satisfying \( f(m) = m \) for all Canadian functions \( f \) *magical*.

4. **Claim 1: \( f(f(x)) = x \):**
   - **Proof:**
     Consider \( P(x, 2x) \):
     \[
     \gcd(f(f(x)), f(2x)) = \gcd(x, x) = x
     \]
     This implies \( x \mid f(f(x)) \) and \( x \mid f(2x) \) for all \( x \).

     Suppose \( p \) is a prime and \( v_p(x) = k \). Assume \( v_p(f(f(x))) > k \) for some \( x \). Take a positive integer \( s \) such that \( sp^{k+1} > x \). Note that \( f(2sp^{k+1}) \geq sp^{k+1} > x \). Since \( x \mid f(2x) \), we have:
     \[
     \gcd(f(f(x)), f(f(2sp^{k+1}))) = \gcd(x, f(2sp^{k+1}))
     \]
     Since \( p^{k+1} \mid f(f(x)) \) and \( p^{k+1} \mid 2sp^{k+1} \mid f(f(2sp^{k+1})) \), the left-hand side (LHS) of the above equation is divisible by \( p^{k+1} \), implying:
     \[
     p^{k+1} \mid \gcd(x, f(2sp^{k+1}))
     \]
     Thus, \( p^{k+1} \mid x \), which is a contradiction. Therefore, \( v_p(x) = v_p(f(f(x))) \) for all primes \( p \), implying \( f(f(x)) = x \) for all \( x \). 

     \(\blacksquare\)

5. **Rewrite \( P(x, y) \) using \( f(f(x)) = x \):**
   \[
   \gcd(x, f(y)) = \gcd(x, y)
   \]
   for all \( x < y \). Let this assertion be \( Q(x, y) \).

6. **Claim 2: \( x \mid f(x) \) for all non-prime powers:**
   - **Proof:**
     Let \( r \) be a prime and \( v_r(x) = \ell \). If \( x \) is not a prime power, assume \( f(x) \neq r^\ell \). Therefore, \( f(x) > r^\ell \). Then, \( P(r^\ell, f(x)) \) implies:
     \[
     \gcd(r^\ell, f(f(x))) = \gcd(r^\ell, f(x))
     \]
     Since \( r^\ell \mid x \mid f(f(x)) \), we have:
     \[
     r^\ell \mid \gcd(r^\ell, f(f(x))) = \gcd(r^\ell, f(x)) \Rightarrow r^\ell \mid f(x)
     \]
     Therefore, \( v_r(f(x)) \geq v_r(x) \) for all primes \( r \), implying \( x \mid f(x) \) for all non-prime powers \( x \).

     \(\blacksquare\)

7. **Claim 3: All non-prime powers are magical:**
   - **Proof:**
     For a non-prime power \( x \), since \( x \mid f(x) \) from Claim 2, \( f(x) \) cannot be a prime power, as otherwise \( x \) would be too. Therefore, \( f(x) \mid f(f(x)) \), resulting in:
     \[
     x \mid f(x) \mid f(f(x)) = x
     \]
     Thus, all non-prime powers are magical.

     \(\blacksquare\)

8. **Prime powers are not magical:**
   Consider the following construction:
   \[
   f(x) = \begin{cases}
   p^\ell & \text{if } x = p^k \\
   p^k & \text{if } x = p^\ell \\
   x & \text{if } x \neq p^k \text{ and } x \neq p^\ell
   \end{cases}
   \]
   where \( p \) is a prime and \( k \neq \ell \). This function is involutive and satisfies \( Q(x, y) \), showing that prime powers are not magical.

The final answer is \(\boxed{\text{all numbers that are not prime powers}}\)