Find all positive integers $b$ with the following property: there exists positive integers $a,k,l$ such that $a^k + b^l$ and $a^l + b^k$ are divisible by $b^{k+l}$ where $k \neq l$.
1. We need to find all positive integers \( b \) such that there exist positive integers \( a, k, l \) with \( k \neq l \) and both \( a^k + b^l \) and \( a^l + b^k \) are divisible by \( b^{k+l} \).

2. Consider the \( p \)-adic valuation \( \nu_p \) for a prime \( p \). Recall that for any integers \( x \) and \( y \), if \( \nu_p(x) \neq \nu_p(y) \), then \( \nu_p(x+y) = \min(\nu_p(x), \nu_p(y)) \).

3. Suppose \( b \) has the desired property and let \( p \) be a prime divisor of \( b \). We analyze the \( p \)-adic valuations of the terms involved.

4. If \( \nu_p(a^k) = k \nu_p(a) \) and \( \nu_p(b^l) = l \nu_p(b) \), then for \( a^k + b^l \) to be divisible by \( b^{k+l} \), we need:
   \[
   \nu_p(a^k + b^l) \geq (k+l) \nu_p(b)
   \]

5. Similarly, for \( a^l + b^k \) to be divisible by \( b^{k+l} \), we need:
   \[
   \nu_p(a^l + b^k) \geq (k+l) \nu_p(b)
   \]

6. Suppose \( k \nu_p(a) = l \nu_p(b) \). Then:
   \[
   \nu_p(a^k) = k \nu_p(a) = l \nu_p(b)
   \]
   and
   \[
   \nu_p(b^l) = l \nu_p(b)
   \]
   Thus, \( \nu_p(a^k) = \nu_p(b^l) \).

7. Similarly, if \( l \nu_p(a) = k \nu_p(b) \), then:
   \[
   \nu_p(a^l) = l \nu_p(a) = \frac{l^2}{k} \nu_p(b)
   \]
   and
   \[
   \nu_p(b^k) = k \nu_p(b)
   \]
   Since \( k \neq l \), \( \frac{l^2}{k} \neq k \), so \( \nu_p(a^l) \neq \nu_p(b^k) \).

8. Therefore, either \( \nu_p(a^k) \neq \nu_p(b^l) \) or \( \nu_p(a^l) \neq \nu_p(b^k) \). This implies:
   \[
   \min(\nu_p(a^k + b^l), \nu_p(a^l + b^k)) \leq \max(\min(\nu_p(a^k), \nu_p(b^l)), \min(\nu_p(a^l), \nu_p(b^k))) \leq \max(\nu_p(b^l), \nu_p(b^k)) < \nu_p(b^{k+l})
   \]

9. Hence, \( b^{k+l} \) cannot divide both \( a^k + b^l \) and \( a^l + b^k \) unless \( b = 1 \).

10. If \( b = 1 \), then \( a^k + 1 \) and \( a^l + 1 \) are trivially divisible by \( 1^{k+l} = 1 \).

Conclusion:
The only possible value for \( b \) is \( 1 \).

The final answer is \( \boxed{1} \)