Consider the question 1: Let $\frac{a_{1}}{b_{1}}$ and $\frac{a_{2}}{b_{2}}$ be the simplest proper fractions, and $a_{2} b_{1}-a_{1} b_{2}=1$. Then the smallest positive integer satisfying $\frac{a_{1}}{b_{1}}<\frac{k}{n}<\frac{a_{2}}{b_{2}}$ is
$$
n=b_{1}+b_{2}, k=a_{1}+a_{2} .
$$
Prove that from $\frac{a_{1}}{b_{1}}a_{1} n, a_{2} n>b_{2} k$.
Therefore, there exist positive integers $A, B$ such that $\left\{\begin{array}{l}b_{1} k-a_{1} n=A, \\ a_{2} n-b_{2} k=B .\end{array}\right.$
Solving the system of equations, we get $\left\{\begin{array}{l}n=\frac{b_{1} B+b_{2} A}{a_{2} b_{1}-a_{1} b_{2}}=b_{1} B+b_{2} A\left(a_{2} b_{1}-a_{1} b_{2}=1\right), \\ k=\frac{a_{1} B+a_{2} A}{a_{2} b_{1}-a_{1} b_{2}}=a_{1} B+a_{2} A .\end{array}\right.$
When $A, B$ both take the minimum value 1, $n, k$ also take the minimum values $n=b_{1}+b_{2}, k=a_{1}+a_{2}$.
Explanation (1) gives the expression for the smallest positive integer solutions of $n, k$. We naturally wonder: Is there a positive integer $k$ that makes $n$ take the maximum value? After research, the following result is obtained:

Let $\frac{a_{1}}{b_{1}}$ and $\frac{a_{2}}{b_{2}}$ be the simplest proper fractions, and $a_{2} b_{1}-a_{1} b_{2}=1$. Then, for $\frac{a_{1}}{b_{1}}<\frac{k}{n}<\frac{a_{2}}{b_{2}}$, there is a unique positive integer $k$ that makes $n$ take the maximum value, which must be $n=2 b_{1} b_{2}, k=a_{1} b_{2}+a_{2} b_{1}$.

Proof: From the above, we know that $n, k$ satisfying $\frac{a_{1}}{b_{1}}<\frac{k}{n}<\frac{a_{2}}{b_{2}}$ are (1). To make $k$ take a unique value and $n$ as large as possible, we transform the above inequality to $\frac{a_{1} b_{2}}{b_{1} b_{2}}<\frac{a_{1} B+a_{2} A}{b_{1} B+b_{2} A}<\frac{a_{2} b_{1}}{b_{1} b_{2}}$.
Since $a_{2} b_{1}-a_{1} b_{2}=1$, we further transform the inequality to $\frac{2 a_{1} b_{2}}{2 b_{1} b_{2}}<\frac{a_{1} B+a_{2} A}{b_{1} B+b_{2} A}<\frac{2 a_{2} b_{1}}{2 b_{1} b_{2}}$,
At this point, $k$ takes the unique value $k=a_{1} B+a_{2} A=2 a_{1} b_{2}+1=2 a_{2} b_{1}-1$.
Substituting $a_{2} b_{1}=a_{1} b_{2}+1$ into the above, we get
$$
k=a_{1} B+a_{2} A=2 a_{1} b_{2}+1=a_{1} b_{2}+\left(a_{1} b_{2}+1\right)=a_{1} b_{2}+a_{2} b_{2} .
$$

That is, taking $B=b_{2}, A=b_{1}$, substituting into (1) gives $n=b_{1} B+b_{2} A=2 b_{1} b_{2}$.
When $n$ increases further, $k$ is no longer unique. Hence, this $n$ is the maximum value.