Given $p$ is a positive integer and a constant. Find the maximum and minimum positive integers $n$, for which there exists a unique positive integer $k$ satisfying
$$
\frac{p+1}{2 p+1}<\frac{n}{n+k}<\frac{p}{2 p-1} .
$$
Solution: Since $p$, $n$, and $k$ are all positive integers, transforming the original inequality yields
$$
\frac{n(p-1)}{p}<k<\frac{n p}{p+1} \text{. }
$$

For $n$, there is a unique positive integer $k$ that satisfies equation (1), so,
$$
\frac{n p}{p+1}-\frac{n(p-1)}{p} \leqslant 2 \text{. }
$$

Solving this, we get $n \leqslant 2 p(p+1)$.
When $n=2 p(p+1)$, substituting into equation (1) gives
$$
2 p^{2}-2<k<2 p^{2}, k=2 p^{2}-1 \text{. }
$$

Thus, the maximum value of $n$ is $2 p(p+1)$.
On the other hand, according to equation (1), from $\frac{n(p-1)}{p}<k$ we get $n(p-1)<k p$.
Therefore, $n(p-1)+1 \leqslant k p$.
Solving this, we get $k \geqslant \frac{n(p-1)+1}{p}$.
From $k<\frac{n p}{p+1}$ we get $k(p+1)<n p$. Therefore,
$$
k(p+1)+1 \leqslant n p \text{. }
$$

Solving this, we get $k \leqslant \frac{n p-1}{p+1}$.
From equations (2) and (3), we get
$$
\frac{n(p-1)+1}{p} \leqslant k \leqslant \frac{n p-1}{p+1} \text{. }
$$

Solving this, we get $n \geqslant 2 p+1$.
When $n=2 p+1$, substituting into equation (4) gives $k=2 p-1$. Therefore, the minimum value of $n$ is $2 p+1$.