The smallest real number $\lambda$ is $\frac{1}{2 n}$.
(1) $\lambda$ can take $\frac{1}{2 n}$.

Let $a_{i}=\frac{2 i-1}{2 n}$.
Below, we show: For any
$$
0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n} \leqslant 1,
$$

there always exists an $i$ such that
$$
\begin{array}{l}
\left|x_{i}-a_{i}\right| \leqslant \frac{1}{2 n} . \\
\text { Let } I_{1}=\left[0, \frac{1}{n}\right], I_{2}=\left(\frac{1}{n}, \frac{2}{n}\right], \cdots, I_{n}=\left(\frac{n-1}{n}, 1\right] .
\end{array}
$$

Define the function $f:\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, n\}$: if $x_{i} \in I_{k_{i}}$, then let $f(i)=k_{i}$.

Below, we prove: $f(x)$ has a fixed point, i.e., there exists an $i \in$ $\{1,2, \cdots, n\}$ such that $f(i)=i$.
Proof by contradiction.
If not, for any $i \in\{1,2, \cdots, n\}$, we have $f(i) \neq i$.
Since $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, it follows that
$$
k_{1} \leqslant k_{2} \leqslant \cdots \leqslant k_{n} \text {. }
$$

And $f(i)=k_{i}$, by assumption, $k_{i} \neq i$.
Then there exists a $j \leqslant n-1$, such that
$$
f(n)=k_{n}=j \leqslant n-1 \text {. }
$$

Thus, for any $i \in\{1,2, \cdots, n-1\}$, we have
$$
f(i) \leqslant j-1 \text {. }
$$

This leads to the existence of a $t$, such that
$$
f(t)=t \text { or } f(1) \leqslant 0 \text {, }
$$

which is a contradiction.
(2) Show that $\lambda \geqslant \frac{1}{2 n}$.
Proof by contradiction.
If not, suppose $a_{1}, a_{2}, \cdots, a_{n}$ and $\lambda^{\prime}<\frac{1}{2 n}$ satisfy the requirement.

Notice that, the union of $n$ closed intervals centered at $a_{1}, a_{2}, \cdots, a_{n}$ with radius $\lambda^{\prime}$ has a total length of no more than $n \cdot 2 \lambda^{\prime}<1$.
Thus, there must exist a point $x^{\prime} \in[0,1]$ that does not fall into any of the above closed intervals.
Taking $x_{1}=x_{2}=\cdots=x_{n}=x^{\prime}$, we reach a contradiction.
In conclusion, the smallest real number $\lambda$ is $\frac{1}{2 n}$.
Given a positive integer $n$. Find the smallest real number $\lambda$, such that there exist real numbers $a_{1}, a_{2}, \cdots, a_{n}$ in the interval $[0,1]$, for any real numbers $x_{i}(i=1,2, \cdots, n)$ satisfying $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n} \leqslant 1$, we have
$$
\min _{1 \leq i \leq n}\left|x_{i}-a_{i}\right| \leqslant \lambda
$$