28. Let $n \in \mathbf{N}^{\cdot}, I$ be an open interval on the number line with length $\frac{1}{n}$. Find the maximum number of simplest fractions $\frac{a}{b}$ such that $1 \leqslant b \leqslant n$ and $\frac{a}{b} \in I$.
28. Let $\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{r}}{b_{r}} \in I$ be all the simplest fractions in $I^{\prime}$ satisfying: $1 \leqslant b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{r} \leqslant n$. Then for $1 \leqslant i\left|\frac{a_{j}}{b_{j}}-\frac{a_{i}}{b_{i}}\right|=\frac{M}{\left[b_{i}, b_{j}\right]}$

Here $M$ is a positive integer. Therefore, $\left[b_{i}, b_{j}\right]>n$. That is, the least common multiple of any two numbers in $b_{1}, b_{2}, \cdots, b_{r}$ is greater than $n$. We will prove: $r \leqslant\left[\frac{n+1}{2}\right]$.

In fact, let the odd numbers from $1,2, \cdots, n$ be $x_{1}, x_{2}, \cdots, x_{m}$ in ascending order, then $m \leqslant\left[\frac{n+1}{2}\right]$. Let $x_{j}=\left\{2^{a} \cdot x_{j} \mid \alpha \in \mathbf{N}\right\}$, and $T_{j}=x_{j} \cap\{1,2, \cdots, n\}$, then $T_{1}$, $T_{2}, \cdots, T_{m}$ form a partition of $\{1,2, \cdots, n\}$, and in $T_{j}$, the larger number is a multiple of the smaller number. If $r>\left[\frac{n+1}{2}\right]$, then there must be two numbers in $b_{1}, b_{2}, \cdots, b_{r}$ that belong to the same $T_{j}$, and the least common multiple of these two numbers is no greater than $n$, which is a contradiction. Therefore, $r \leqslant\left[\frac{n+1}{2}\right]$.

On the other hand, when $n=2 p+1$, for any $j \in\{1,2, \cdots, p+1\}$, we have $\frac{1}{p+j} \in \left(\frac{1}{n}-\varepsilon, \frac{2}{n}-\varepsilon\right)$, where $\varepsilon=\frac{1}{2(2 p+1)(p+1)}$; and when $n=2 p$, for any $j \in \{1,2, \cdots, p\}$, we have $\frac{1}{p+j} \in\left(\frac{1}{n}-\varepsilon, \frac{2}{n}-\varepsilon\right)$, where $\varepsilon=\frac{1}{2 p(p+1)}$. Therefore, some $I$ can contain $\left[\frac{n+1}{2}\right]$ numbers.

In summary, the maximum value sought is $\left[\frac{n+1}{2}\right]$.