Let $p, q$ be coprime positive integers. A set $S=\left\{a_{1}, a_{2}, \cdots, a_{p+q}\right\}\left(a_{1}<a_{2}<\cdots<a_{p+q}\right)$ of $p+q$ real numbers is called balanced if $a_{1}, a_{2}, \cdots, a_{p}$ form an arithmetic sequence with common difference $q$, and $a_{p}, a_{p+1}, \cdots, a_{p+q}$ form an arithmetic sequence with common difference $p$. Suppose there are $k$ balanced sets of $p+q$ elements, and the intersection of any two of them is non-empty. Find the maximum possible value of $k$.
The maximum value of the required $k$ is $p+\max \{p, q\}$.
Note that the intersection of two balanced sets is non-empty, which is equivalent to one set being obtained from the other through a translation transformation $k q+l q$ $(k \in\{0,1, \cdots, q\}, l \in\{0,1, \cdots, p-1\})$. Therefore, the maximum value sought is the maximum number of integers of the form $k p+l q(k \in\{0,1, \cdots, q\}, l \in\{0,1, \cdots, p-1\})$.

We will prove that there are at most $p+\max \{p, q\}$ such integers. Let $r=\max \{p, q\}$. Suppose there are $p+r+1$ such integers, which we arrange in lexicographical order as $k_{i} p+l_{i} q(i=1,2, \cdots, p+r+1)$, where $k_{i} \leqslant k_{i+1}$, and if $k_{i}=k_{i+1}$, then $l_{i} \leqslant l_{i+1}$. In fact, note that there are at most $q$ indices $i$ such that $k_{i} < k_{i+1}$, and at most $p$ indices $i$ such that $k_{i}=k_{i+1}$ and $l_{i} < l_{i+1}$. Therefore, there are at least $p+r+1 - (q + p) = r - q + 1$ indices $i$ such that $k_{i}=k_{i+1}$ and $l_{i} = l_{i+1}$. This implies that $\left|\left(k_{i} p+l_{i} q\right)-\left(k_{i+1} p+l_{i+1} q\right)\right|=k p+l q$, where $k \in\{0,1, \cdots, q\}, l \in\{0,1, \cdots, p-1\}$. That is, $\left(k_{i}-k_{i+1}\right) p-\left(l_{i}-l_{i+1}\right) q=k p+l q$ or $\left(k_{i+1}-k_{i}\right) p+\left(l_{i+1}-l_{i}\right) q=k p+l q$. By the fact that $p, q$ are coprime, the first equation implies $l_{i}-l_{i+1} \equiv l(\bmod p)$, and the second equation implies $k_{i+1}-k_{i} \equiv k(\bmod q)$. Combining the ranges of $k, l$, the first congruence is actually an equality, leading to $k_{i}-k_{i+1}=k$, and since $0 > k_{i}-k_{i+1}=k \geqslant 0$, this is a contradiction. The second congruence implies $k_{i+1}=q, k_{i}=k=0$ (otherwise, a similar contradiction arises), hence
$$
k_{j}=\left\{\begin{array}{ll}
0, & j=1,2, \cdots, i, \\
q, & j=i+1, i+2, \cdots, p+r .
\end{array}\right.
$$

This indicates that there are at least $\frac{p+r-1}{2}$ indices $j$ such that $k_{j}=k_{j+1}$. In this case, the corresponding indices $j$ satisfy $l_{j} < l_{j+1}$. If $k_{j} < k_{j+1}$, then $k_{j}$ are $\underbrace{0,0, \cdots, 0}_{p \uparrow}, \underbrace{q, q, \cdots, q}_{p \uparrow}$, and the corresponding $l_{j}$ are $0,1,2, \cdots, p-1,0,1,2, \cdots, p-1$. It is easy to verify that these numbers satisfy the conditions.