For postive integers $k,n$, let
$$f_k(n)=\sum_{m\mid n,m>0}m^k$$
Find all pairs of positive integer $(a,b)$ such that $f_a(n)\mid f_b(n)$ for every positive integer $n$. 
1. **Consider the function \( f_k(n) \):**
   \[
   f_k(n) = \sum_{m \mid n, m > 0} m^k
   \]
   This function sums the \( k \)-th powers of all positive divisors of \( n \).

2. **Simplify the problem by considering \( n \) as a prime power:**
   Let \( n = p^k \) where \( p \) is a prime and \( k \) is a positive integer. We need to find pairs \((a, b)\) such that \( f_a(n) \mid f_b(n) \) for every positive integer \( n \).

3. **Evaluate \( f_a(p^k) \) and \( f_b(p^k) \):**
   \[
   f_a(p^k) = \sum_{m \mid p^k} m^a = 1^a + p^a + p^{2a} + \cdots + p^{ka} = \frac{p^{(k+1)a} - 1}{p^a - 1}
   \]
   \[
   f_b(p^k) = \sum_{m \mid p^k} m^b = 1^b + p^b + p^{2b} + \cdots + p^{kb} = \frac{p^{(k+1)b} - 1}{p^b - 1}
   \]

4. **Condition for divisibility:**
   We need \( f_a(p^k) \mid f_b(p^k) \). This implies:
   \[
   \frac{p^{(k+1)a} - 1}{p^a - 1} \mid \frac{p^{(k+1)b} - 1}{p^b - 1}
   \]

5. **Simplify the divisibility condition:**
   For \( f_a(p^k) \mid f_b(p^k) \) to hold, the numerator of \( f_b(p^k) \) must be divisible by the numerator of \( f_a(p^k) \) and the denominator of \( f_a(p^k) \) must divide the denominator of \( f_b(p^k) \).

6. **Analyze the condition for prime powers:**
   Consider the case when \( k = 1 \):
   \[
   f_a(p) = 1 + p^a
   \]
   \[
   f_b(p) = 1 + p^b
   \]
   For \( 1 + p^a \mid 1 + p^b \), it must be that \( b = ta \) for some integer \( t \).

7. **Generalize the condition:**
   For any \( k \), we have:
   \[
   f_a(p^k) = \frac{p^{(k+1)a} - 1}{p^a - 1}
   \]
   \[
   f_b(p^k) = \frac{p^{(k+1)b} - 1}{p^b - 1}
   \]
   For \( f_a(p^k) \mid f_b(p^k) \), it must be that \( b = ta \) for some integer \( t \).

8. **Consider the case when \( t \geq 3 \):**
   If \( t \geq 3 \), i.e., \( b > a \), choose \( p = 5 \) and let \( r \) be a prime number that divides \( p^b - 1 \) but does not divide \( p^t - 1 \) for any \( 1 \leq t < b \). Also, choose \( k + 1 = b \).

9. **Evaluate the \( \nu_r \) function:**
   \[
   \nu_r(f_a(p^k)) = \nu_r(p^{\ell} - 1) + \nu_r\left( \frac{(k+1)a}{\ell} \right) = \nu_r(p^b - 1) + \nu_r(a)
   \]
   \[
   \nu_r(f_b(p^k)) = \nu_r(k + 1) = \nu_r(b)
   \]
   Since \( r \mid p^{r-1} - 1 \), we get \( b \leq r - 1 \) implying \( \nu_r(a) = \nu_r(b) = 0 \).

10. **Contradiction:**
    \[
    \nu_r(f_a(p^k)) < \nu_r(f_b(p^k))
    \]
    This leads to a contradiction.

11. **Conclusion:**
    Hence, the only solution is \( (a, b) = (c, c) \) for every positive integer \( c \).

The final answer is \( \boxed{ (a, b) = (c, c) } \) for every positive integer \( c \).