Let $p,q$ be positive integers. For any $a,b\in\mathbb{R}$ define the sets $$P(a)=\bigg\{a_n=a \ + \ n \ \cdot \ \frac{1}{p} : n\in\mathbb{N}\bigg\}\text{ and }Q(b)=\bigg\{b_n=b \ + \ n \ \cdot \ \frac{1}{q} : n\in\mathbb{N}\bigg\}.$$
The [i]distance[/i] between $P(a)$ and $Q(b)$ is the minimum value of $|x-y|$ as $x\in P(a), y\in Q(b)$. Find the maximum value of the distance between $P(a)$ and $Q(b)$ as $a,b\in\mathbb{R}$.
1. **Define the sets \( P(a) \) and \( Q(b) \):**
   \[
   P(a) = \left\{ a_n = a + n \cdot \frac{1}{p} : n \in \mathbb{N} \right\}
   \]
   \[
   Q(b) = \left\{ b_n = b + n \cdot \frac{1}{q} : n \in \mathbb{N} \right\}
   \]

2. **Define the distance between \( P(a) \) and \( Q(b) \):**
   \[
   \text{dist}(P(a), Q(b)) = \min_{x \in P(a), y \in Q(b)} |x - y|
   \]
   This can be rewritten as:
   \[
   \text{dist}(P(a), Q(b)) = \min_{m, n \in \mathbb{N}} \left| a - b + \frac{m}{p} - \frac{n}{q} \right|
   \]

3. **Extend \( m, n \) to \( \mathbb{Z} \):**
   \[
   \text{dist}(P(a), Q(b)) = \min_{m, n \in \mathbb{Z}} \left| a - b + \frac{m}{p} - \frac{n}{q} \right|
   \]

4. **Express \( \frac{m}{p} - \frac{n}{q} \) in terms of the least common multiple (LCM):**
   \[
   \frac{m}{p} - \frac{n}{q} = \frac{mq - np}{pq}
   \]
   Since \( \text{lcm}(p, q) = \frac{pq}{\gcd(p, q)} \), we can write:
   \[
   \frac{mq - np}{pq} = \frac{\ell \cdot \gcd(p, q)}{pq} = \frac{\ell}{\text{lcm}(p, q)}
   \]
   where \( \ell \in \mathbb{Z} \) by Bezout's Theorem.

5. **Rewrite the distance in terms of \( \ell \):**
   \[
   \text{dist}(P(a), Q(b)) = \min_{\ell \in \mathbb{Z}} \left| a - b + \frac{\ell}{\text{lcm}(p, q)} \right|
   \]

6. **Analyze the maximum distance:**
   Suppose there exists \( a, b \in \mathbb{R} \) such that:
   \[
   \text{dist}(P(a), Q(b)) > \frac{1}{2 \cdot \text{lcm}(p, q)}
   \]
   By definition:
   \[
   \min_{\ell \in \mathbb{Z}} \left| a - b + \frac{\ell}{\text{lcm}(p, q)} \right| > \frac{1}{2 \cdot \text{lcm}(p, q)}
   \]
   If \( a > b \), replace \( \ell \) with \( \ell - 1 \) to attain a contradiction. If \( a < b \), replace \( \ell \) with \( \ell + 1 \) to attain a contradiction. Hence:
   \[
   \text{dist}(P(a), Q(b)) \le \frac{1}{2 \cdot \text{lcm}(p, q)}
   \]

7. **Verify equality:**
   For equality, check that \( a = \frac{1}{2 \cdot \text{lcm}(p, q)}, b = 0 \) works:
   \[
   \min_{\ell \in \mathbb{Z}} \left| \frac{1}{2 \cdot \text{lcm}(p, q)} + \frac{\ell}{\text{lcm}(p, q)} \right| = \min_{k \in \mathbb{Z}} \left| \frac{2k + 1}{2 \cdot \text{lcm}(p, q)} \right| \ge \frac{1}{2 \cdot \text{lcm}(p, q)}
   \]

The final answer is \(\boxed{\frac{1}{2 \cdot \text{lcm}(p, q)}}\)