Find all pairs $(a,b)$ of positive integers such that $a^{2017}+b$ is a multiple of $ab$.
1. We need to find all pairs \((a, b)\) of positive integers such that \(ab \mid a^{2017} + b\).

2. We start by rewriting the condition \(ab \mid a^{2017} + b\) as:
   \[
   a^{2017} + b = k \cdot ab \quad \text{for some integer } k.
   \]
   Rearranging, we get:
   \[
   a^{2017} + b = k \cdot ab \implies b(1 - k \cdot a) = -a^{2017}.
   \]

3. We claim that \(a^{2017} \mid b\). Suppose there exists a prime \(p\) such that \(\nu_p(a^{2017}) > \nu_p(b)\). Here, \(\nu_p(n)\) denotes the p-adic valuation of \(n\), i.e., the highest power of \(p\) dividing \(n\).

4. We have:
   \[
   \nu_p(ab) = \nu_p(a) + \nu_p(b) = \frac{\nu_p(a^{2017})}{2017} + \nu_p(b) > \nu_p(b),
   \]
   and:
   \[
   \nu_p(a^{2017} + b) = \nu_p(b).
   \]
   This leads to a contradiction because \(\nu_p(ab)\) should be less than or equal to \(\nu_p(a^{2017} + b)\). Therefore, our claim that \(a^{2017} \mid b\) must be true.

5. Let \(b = k \cdot a^{2017}\) for some integer \(k\). Substituting \(b\) in the original condition, we get:
   \[
   ab \mid a^{2017} + b \implies a \cdot (k \cdot a^{2017}) \mid a^{2017} + k \cdot a^{2017}.
   \]
   Simplifying, we have:
   \[
   k \cdot a^{2018} \mid (k + 1) \cdot a^{2017} \implies k \cdot a \mid k + 1.
   \]

6. We now analyze the divisibility condition \(k \cdot a \mid k + 1\):
   - If \(a = 1\), then \(k \mid k + 1\). This implies \(k = 1\) because the only divisor of \(k + 1\) that is less than or equal to \(k\) is 1. Thus, \((a, b) = (1, 1)\) is a solution.
   - If \(a = 2\), then \(2k \mid k + 1\). This implies \(2k \le k + 1\), which simplifies to \(k \le 1\). Since \(k\) is a positive integer, \(k = 1\). Thus, \((a, b) = (2, 2^{2017})\) is a solution.
   - If \(a > 2\), then \(k \cdot a \le k + 1\). This implies \(k \cdot a - k \le 1\), or \(k(a - 1) \le 1\). Since \(a > 2\), \(a - 1 \ge 2\), and thus \(k(a - 1) \ge 2k\). For \(k(a - 1) \le 1\) to hold, \(k\) must be 0 or 1. Since \(k\) is a positive integer, \(k = 1\). However, if \(k = 1\), then \(a \mid 2\), which is not possible for \(a > 2\).

7. Therefore, the only solutions are \((a, b) = (1, 1)\) and \((a, b) = (2, 2^{2017})\).

The final answer is \(\boxed{(1,1), (2,2^{2017})}\).