Find all pairs of positive integers $(s, t)$, so that for any two different positive integers $a$ and $b$ there exists some positive integer $n$, for which $$a^s + b^t | a^n + b^{n+1}.$$

[i]Proposed by Oleksii Masalitin (Ukraine)[/i]
To solve the problem, we need to find all pairs of positive integers \((s, t)\) such that for any two different positive integers \(a\) and \(b\), there exists some positive integer \(n\) for which \(a^s + b^t \mid a^n + b^{n+1}\).

1. **Case \(d = |s - t| \geq 2\):**
   - Consider \(d = |s - t|\). If \(d \geq 2\), we can find an odd prime divisor \(p\) of \(7^d + 1\). Note that \(7^d + 1\) is not divisible by \(16\) when \(d \geq 2\), and it is larger than \(16\), so it is not a power of \(2\). Hence, such a prime \(p\) exists.
   - Set \(a = 7\) and \(b = p + 7\). Then, \(a^s + b^t \equiv 7^s + 7^t \equiv 0 \pmod{p}\), implying \(p \mid a^n + b^{n+1} \equiv 7^n \cdot 8 \pmod{p}\).
   - Since \(p \neq 7\), \(7^n \cdot 8 \equiv 0 \pmod{p}\) is impossible. Therefore, \(d \geq 2\) does not work.

2. **Case \(d = 0\):**
   - If \(d = 0\), then \(s = t\). Take coprime positive integers \(a\) and \(b\) greater than \(1\). We need \(a^s + b^s \mid a^n + b^{n+1}\).
   - Let \(n = ks + r\) where \(0 \leq r \leq s - 1\). Then, \(a^{ks + r} \equiv (a^s)^k \cdot a^r \equiv (-b^s)^k \cdot a^r \pmod{a^s + b^s}\).
   - Since \(a\) and \(b\) are coprime, we can cancel out \(b^{ks}\) and get \(a^s + b^s \mid (-1)^k \cdot a^r + b^{r+1}\).
   - This implies \((-1)^k \cdot a^r + b^{r+1} < a^s + b^s\) and \((-1)^k \cdot a^r + b^{r+1} > -(a^s + b^s)\). Since \(a\) and \(b\) are coprime and \(b > 1\), this leads to a contradiction.

3. **Case \(d = 1\):**
   - Finally, we show that \(d = 1\) works. If \(t = s + 1\), just take \(n = s\).
   - If \(t = s - 1\), we need \(a^{t+1} + b^t \mid a^n + b^{n+1}\). Consider the prime divisors of \(a^{t+1} + b^t\).
   - Let \(p^r \mid \mid a^{t+1} + b^t\). If \(p \mid a, b\), for all \(n \geq r\), \(p^r \mid a^n + b^{n+1}\).
   - Suppose \(p\) is coprime to \(a\) and \(b\). Since \(1 \equiv a^{\varphi(p^r)} \equiv b^{\varphi(p^r)} \pmod{p^r}\), for all integers \(l\), \(a^{l \varphi(p^r) - t - 1} \equiv -b^{l \varphi(p^r) - t} \pmod{p^r}\).
   - Thus, for that particular prime divisor, all \(n\) such that \(\varphi(p^r) \mid n + t + 1\) satisfy the condition.
   - We can choose \(n = u \varphi(a^{t+1} + b^t) - t - 1\) for \(u\) sufficiently large to make \(n\) larger than all the exponents of prime divisors of \(a^{t+1} + b^t\).

Therefore, the pairs \((s, t)\) that work are those where \(s\) and \(t\) differ by 1.

The final answer is \( \boxed{ (s, t) } \) such that \(|s - t| = 1\).