The 28th question: Let $n(n \geq 2)$ be a given integer. Find the smallest real number $\lambda$ such that for any real numbers $x_{1}, x_{2}, \ldots, x_{n} \in [0,1]$, there exist $\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{n} \in \{0,1\}$, satisfying: for any $1 \leq i \leq j \leq n$, we have $\left|\sum_{k=i}^{j}(\varepsilon_{k} - x_{k})\right| \leq \lambda_{0}$
Question 28:
Solution: (1) We first prove: $\lambda \geq \frac{n}{n+1}$.
Take $\mathrm{x}_{1}=\mathrm{x}_{2}=\ldots=\mathrm{x}_{\mathrm{n}}=\frac{1}{\mathrm{n}+1}$. We only need to prove: For any $\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{\mathrm{n}} \in\{0,1\}$, there exists $1 \leq \mathrm{i} \leq \mathrm{j} \leq \mathrm{n}$, such that $\left|\sum_{k=i}^{j}\left(\varepsilon_{k}-x_{k}\right)\right| \geq \frac{n}{n+1}$.

In fact, if there exists some $\varepsilon_{t}=1$, then $\left|\varepsilon_{k}-x_{k}\right|=\frac{n}{n+1}$; if for any $1 \leq i \leq n$, we have $\varepsilon_{i}=0$, then $\left|\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\varepsilon_{\mathrm{k}}-\mathrm{x}_{\mathrm{k}}\right)\right|=\frac{\mathrm{n}}{\mathrm{n}+1}$. Therefore, $\lambda \geq \frac{\mathrm{n}}{\mathrm{n}+1}$.
Hence, $\lambda \geq \frac{\mathrm{n}}{\mathrm{n}+1}$.
(2) We then prove: $\lambda=\frac{\mathrm{n}}{\mathrm{n}+1}$ satisfies the condition.
For this, we only need to prove: For any $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{\mathrm{n}} \in[0,1]$, there exists a set of $\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{\mathrm{n}} \in \{0,1\}$, such that for any $1 \leq i \leq j \leq n$, we have $\left|\sum_{k=i}^{j}\left(\varepsilon_{k}-x_{k}\right)\right| \leq \frac{n}{n+1}$.
$$
\text{Let } S_{0}=0, S_{k}=\sum_{i=1}^{k} x_{i}(1 \leq k \leq n).
$$

Divide the interval $[0,1)$ into $n+1$ subintervals: $I_{k}=\left[\frac{k}{n+1}, \frac{k+1}{n+1}\right)(0 \leq k \leq n)$. According to the pigeonhole principle, there must exist some subinterval that does not contain $\left\{S_{1}\right\}, \left\{S_{2}\right\}, \cdots, \left\{S_{n}\right\}$, let this interval be $I_{t}=\left[\frac{t}{n+1}, \frac{t+1}{n+1}\right)$. Here $\{x\}$ denotes the fractional part of $x$.

Below is the method for choosing $\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{\mathrm{n}}$: For any $1 \leq \mathrm{k} \leq \mathrm{n}$, if there exists a non-negative integer $\mathrm{m}$, such that $\mathrm{S}_{\mathrm{k}-1} \leq \mathrm{m}+\frac{\mathrm{t}}{\mathrm{n}+1}<\mathrm{m}+\frac{\mathrm{t}+1}{\mathrm{n}+1} \leq \mathrm{S}_{\mathrm{k}}$, then take $\varepsilon_{\mathrm{k}}=1$; otherwise take $\varepsilon_{\mathrm{k}}=0$.
Next, we verify that this choice satisfies the condition, i.e., for any $1 \leq i \leq j \leq n$, we have $\left|\sum_{k=i}^{j}\left(\varepsilon_{k}-x_{k}\right)\right| \leq \frac{n}{n+1}$.
In fact, according to the choice of $\mathrm{I}_{\mathrm{t}}$, we know there exist non-negative integers $\mathrm{u}, \mathrm{v}$, satisfying:
$$
\begin{array}{l}
u-1+\frac{t+1}{n+1} \leq S_{i-1} \leq u+\frac{t}{n+1} \\
v-1+\frac{t+1}{n+1} \leq S_{j} \leq v+\frac{t}{n+1}
\end{array}
$$

Thus, we know
$$
\sum_{k=i}^{j} x_{k}=S_{j}-S_{i-1} \in\left[v-u-\frac{n}{n+1}, v-u+\frac{n}{n+1}\right]
$$

By the choice of $\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{\mathrm{n}}$, we know that among $\varepsilon_{\mathrm{i}}, \varepsilon_{\mathrm{i}+1}, \ldots, \varepsilon_{\mathrm{j}}$, there are exactly $\mathrm{v}-\mathrm{u}$ ones, hence $\sum_{\mathrm{k}=\mathrm{i}}^{\mathrm{j}} \varepsilon_{\mathrm{k}}=\mathrm{v}-u$. Therefore, we know:
$$
\begin{array}{l}
\sum_{k=1}^{j}\left(\varepsilon_{k}-x_{k}\right) \in\left[-\frac{n}{n+1}, \frac{n}{n+1}\right] \\
\Rightarrow\left|\sum_{k=i}^{j}\left(\varepsilon_{k}-x_{k}\right)\right| \leq \frac{n}{n+1}
\end{array}
$$

In summary, the proposition is proved!