To find all pairs of prime numbers $(p, q)$ such that $pq \mid 5^p + 5^q$, we consider several cases:

1. **Case 1: Either $p$ or $q$ is 2.**
- Assume $p = 2$. Then $2q \mid 5^2 + 5^q$.
- This simplifies to $2q \mid 25 + 5^q$.
- Since $25 \equiv 1 \pmod{2}$, we have $5^q + 1 \equiv 0 \pmod{q}$, implying $q \mid 30$.
- The prime divisors of 30 are 2, 3, and 5. Since $q \neq 2$, the possible values for $q$ are 3 and 5.
- Thus, the pairs $(p, q)$ are $(2, 3)$ and $(2, 5)$.

2. **Case 2: $p = q = 5$.**
- Clearly, $5^2 \mid 5^5 + 5^5$, so $(5, 5)$ is a solution.

3. **Case 3: $p = 5$ and $q \neq 5$.**
- Then $5q \mid 5^5 + 5^q$.
- This simplifies to $5q \mid 3125 + 5^q$.
- Since $3125 \equiv 0 \pmod{5}$, we have $5^q + 3125 \equiv 0 \pmod{q}$, implying $q \mid 3126$.
- The prime divisors of 3126 are 2 and 313. Thus, the pairs $(p, q)$ are $(5, 2)$ and $(5, 313)$.

4. **Case 4: $p \neq 5$ and $q \neq 5$.**
- Assume $p > q$. Then $pq \mid 5^p + 5^q$.
- This implies $5^q(5^{p-q} + 1) \equiv 0 \pmod{pq}$.
- Therefore, $5^{p-q} \equiv -1 \pmod{q}$ and $\text{ord}_q(5) \mid 2(p-q)$.
- Since $\text{ord}_q(5) \nmid p-q$, we get a contradiction in the valuation of 2, leading to no further solutions.

Combining all cases, the pairs of prime numbers $(p, q)$ that satisfy the condition are:
$$\boxed{(2, 3), (2, 5), (3, 2), (5, 2), (5, 5), (5, 313), (313, 5)}$$