Example 1 Given a positive integer $n(n>3)$, let real numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n} \geqslant n, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2} .
\end{array}
$$

Find the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
(28th United States of America Mathematical Olympiad)
The above equation holds for $a \geqslant 2$.
Try taking $a=2$, at this point, $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ $=2$. From this, we can conjecture that the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is 2.

Solution: First, we prove that for any real numbers $a_{1}, a_{2}, \cdots, a_{n} (n>3)$ satisfying $a_{1}+a_{2}+\cdots +a_{n} \geqslant n$ and $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2}$, we have
$\max \left\{\ddot{a}_{1}, a_{2}, \cdots, a_{n}\right\} \geqslant 2$.
We use proof by contradiction.
Assume $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\} < 2, i \vee j \geqslant 0, i+j=n$. Then, $\max \left\{x_{1}, x_{2}, \cdots, x_{i}\right\} < 2$,
\[
\begin{array}{l}
2 i=2+2+\cdots+2 > x_{1}+x_{2}+\cdots+x_{i} \\
\geqslant n+y_{1}+y_{2}+\cdots+y_{j} \\
=i+j+y_{1}+y_{2}+\cdots+y_{j}.
\end{array}
\]

Rearranging gives $i-j > y_{1}+y_{2}+\cdots+y_{j}$.
Since $x_{1}^{2}+x_{2}^{2}+\cdots+x_{i}^{2}+\left(-y_{1}\right)^{2}+\left(-y_{2}\right)^{2}+\cdots+\left(-y_{j}\right)^{2} \geqslant n^{2}$, we have
\[
\begin{array}{l}
4 i > x_{1}^{2}+x_{2}^{2}+\cdots+x_{i}^{2} \\
\geqslant n^{2}-\left(y_{1}^{2}+y_{2}^{2}+\cdots+y_{j}^{2}\right) \\
\geqslant n^{2}-\left(y_{1}+y_{2}+\cdots+y_{j}\right)^{2} \\
> n^{2}-(i-j)^{2} \\
= (i+j)^{2}-(i-j)^{2} = 4 i j.
\end{array}
\]

Since $i \geqslant 0$, we have $j a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2}$.
This leads to a contradiction with $n>3$.
Next, take $a_{1}=a_{2}=\cdots=a_{n}=2$, then $a_{1}+a_{2}+\cdots+a_{n} \geqslant n$ and $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2}$ both hold. At this point,
$\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=2$.
In summary, the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is 2.