To solve the problem, we need to find all prime numbers $p$ and $q$ such that $p$ divides $30q - 1$ and $q$ divides $30p - 1$.

1. Initial Setup:
Given the conditions:
$$p \mid (30q - 1) \quad \text{and} \quad q \mid (30p - 1)$$
This implies there exist integers $a$ and $b$ such that:
$$30q - 1 = ap \quad \text{and} \quad 30p - 1 = bq$$

2. Combining the Equations:
We can rewrite the equations as:
$$30q = ap + 1 \quad \text{and} \quad 30p = bq + 1$$
Adding these two equations, we get:
$$30q + 30p = ap + bq + 2$$
Rearranging, we have:
$$30(p + q) = ap + bq + 2$$

3. **Exploring Possible Values of $k$:**
Let $k = \frac{30(p + q) - 2}{pq}$. Since $k$ must be an integer, we need to find values of $k$ such that:
$$k(pq) = 30(p + q) - 2$$
Given that $k$ must be a divisor of $30$ and $\gcd(30, k) = 1$, the possible values of $k$ are $1, 7, 11, 13, 17, 19, 23, 29$.

4. **Analyzing Each $k$:**
- For $k = 1$:
$$pq = 30(p + q) - 2$$
Rearranging, we get:
$$pq - 30p - 30q + 2 = 0$$
Solving this quadratic equation in terms of $p$ and $q$, we find:
$$(p - 30)(q - 30) = 899$$
The factor pairs of 899 are $(1, 899)$ and $(29, 31)$. Since $p$ and $q$ are primes, the valid pairs are:
$$(p, q) = (59, 61) \quad \text{and} \quad (61, 59)$$

- For $k = 7$:
$$7pq = 30(p + q) - 2$$
Rearranging, we get:
$$(7p - 30)(7q - 30) = 893$$
The factor pairs of 893 are $(1, 893)$ and $(19, 47)$. Since $p$ and $q$ are primes, the valid pairs are:
$$(p, q) = (7, 11) \quad \text{and} \quad (11, 7)$$

- For $k = 11$:
$$11pq = 30(p + q) - 2$$
Rearranging, we get:
$$(11p - 30)(11q - 30) = 889$$
The factor pairs of 889 are $(1, 889)$ and $(23, 37)$. Since $p$ and $q$ are primes, there are no valid pairs.

5. Conclusion:
The valid pairs of primes $(p, q)$ that satisfy the given conditions are:
$$(p, q) = (7, 11), (11, 7), (59, 61), (61, 59)$$

The final answer is $\boxed{(7, 11), (11, 7), (59, 61), (61, 59)}$.