The 72nd problem: Let $n \geq 2$ be a given positive integer. Find the largest $\lambda=\lambda(n)$ such that $\lambda \cdot \max _{1 \leq k \leq n}\left|z_{k}\right|^{2} \leq$ $\sum_{\mathrm{k}=1}^{\mathrm{n}}\left|\mathrm{z}_{\mathrm{k}+1}-\mathrm{z}_{\mathrm{k}}\right|^{2}$ holds for any complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2}, \ldots, \mathrm{z}_{\mathrm{n}}$ satisfying $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{z}_{\mathrm{k}}=0$, where $\mathrm{z}_{\mathrm{n}+1}=\mathrm{z}_{1}$.
The 72nd problem:

Proof Method One: We first prove a lemma: Let $\mathrm{n} \geq 2$ be a given positive integer, then for any real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfying $\sum_{k=1}^{n} x_{k}=0$, we have $\frac{12 n}{n^{2}-1} \cdot \max _{1 \leq k \leq n} x_{k}^{2} \leq \sum_{k=1}^{n}\left(x_{k+1}-x_{k}\right)^{2}$, where $\mathrm{x}_{\mathrm{n}+1}=\mathrm{x}_{1}$.

Proof of the Lemma: By symmetry, we can assume $\mathrm{x}_{1}=\mathrm{x}_{\mathrm{n}+1}=\max _{1 \leq \mathrm{k} \leq \mathrm{n}}\left|\mathrm{x}_{\mathrm{k}}\right|$. Then, according to the condition, for any real number $c$, we have $n x_{n+1}=\sum_{i=1}^{n}(i+c)\left(x_{i+1}-x_{i}\right)$. Hence, by the Cauchy-Schwarz inequality, we know:
$$
\begin{array}{l}
x_{n+1}^{2}=\frac{1}{n^{2}}\left[\sum_{i=1}^{n}(i+c)\left(x_{i+1}-x_{i}\right)\right]^{2} \leq \frac{1}{n^{2}}\left[\sum_{i=1}^{n}(i+c)^{2}\right]\left[\sum_{k=1}^{n}\left(x_{k+1}-x_{k}\right)^{2}\right] \\
=\frac{1}{n^{2}}\left[n c^{2}+n(n+1) c+\frac{n(n+1)(2 n+1)}{6}\right]\left[\sum_{k=1}^{n}\left(x_{k+1}-x_{k}\right)^{2}\right]
\end{array}
$$

In (1), taking $c=-\frac{\mathrm{n}+1}{2}$, we get:
$$
\begin{array}{l}
x_{n+1}^{2} \leq \frac{1}{n^{2}} \cdot \frac{n\left(n^{2}-1\right)}{12}\left[\sum_{k=1}^{n}\left(x_{k+1}-x_{k}\right)^{2}\right] \\
\Rightarrow \frac{12 n}{n^{2}-1} \cdot x_{n+1}^{2} \leq \sum_{k=1}^{n}\left(x_{k+1}-x_{k}\right)^{2}
\end{array}
$$

The lemma is proved!
Now, let's return to the original problem.
Let $\mathrm{z}_{\mathrm{k}}=\mathrm{x}_{\mathrm{k}}+\mathrm{y}_{\mathrm{k}} \mathrm{i}(1 \leq \mathrm{k} \leq 0)$, then $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{k}}=0$, and $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{y}_{\mathrm{k}}=0$. According to the lemma, we have:
$\frac{12 \mathrm{n}}{\mathrm{n}^{2}-1} \cdot \max _{1 \leq \mathrm{k} \leq \mathrm{n}} \mathrm{x}_{\mathrm{k}}^{2} \leq \sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\mathrm{x}_{\mathrm{k}+1}-\mathrm{x}_{\mathrm{k}}\right)^{2}$
$\frac{12 n}{n^{2}-1} \cdot \max _{1 \leq k \leq n} y_{k}^{2} \leq \sum_{k=1}^{n}\left(y_{k+1}-y_{k}\right)^{2}$
Assume $\mathrm{z}_{\mathrm{n}}=\max _{1 \leq \mathrm{k} \leq \mathrm{n}}\left|\mathrm{z}_{\mathrm{k}}\right|$, then
$$
\begin{array}{l}
\frac{12 n}{n^{2}-1} \max _{1 \leq k \leq n}\left|z_{k}\right|^{2}=\frac{12 n}{n^{2}-1} \cdot\left|z_{n}\right|^{2}=\frac{12 n}{n^{2}-1} \cdot\left(x_{n}^{2}+y_{n}^{2}\right) \\
\leq \frac{12 n}{n^{2}-1} \cdot \max _{1 \leq k \leq n} x_{k}^{2}+\frac{12 n}{n^{2}-1} \cdot \max _{1 \leq k \leq n} y_{k}^{2} \\
\leq \sum_{k=1}^{n}\left(x_{k+1}^{2}-x_{k}\right)^{2}+\sum_{k=1}^{n}\left(y_{k+1}-y_{k}\right)^{2}=\sum_{k=1}^{n}\left|z_{k+1}-z_{k}\right|^{2}
\end{array}
$$

On the other hand, taking $z_{k}=\frac{n^{2}-1}{12}-\frac{(n+1-k)(k-1)}{2}(1 \leq k \leq n)$, the equality holds.
In summary, $\lambda(n)=\frac{12 \mathrm{n}}{\mathrm{n}^{2}-1}$.
Proof Method Two: Let $\max _{1 \leq k \leq n}\left|z_{k}\right|=\left|z_{r}\right|>0$, then we have the identity:
$$
\begin{array}{l}
\frac{1}{n^{2}\left|z_{r}\right|^{2}}\left(\sum_{k=1}^{n}\left|z_{k+1}-z_{k}\right|^{2}\right)-\frac{12}{n\left(n^{2}-1\right)} \\
=\sum_{k=1}^{r}\left|\frac{z_{k+1}-z_{k}}{n z_{r}}-\frac{12(n+k-r)-6(n-1)}{n\left(n^{2}-1\right)}\right|^{2}+\sum_{k=r}^{n}\left|\frac{z_{k+1}-z_{k}}{n z_{r}}-\frac{12(k-r)-6(n-1)}{n\left(n^{2}-1\right)}\right|^{2} \geq 0
\end{array}
$$

Hence, $\frac{12 n}{n^{2}-1} \max _{1 \leq k \leq n}\left|z_{k}\right|^{2} \leq \sum_{k=1}^{n}\left|z_{k+1}-z_{k}\right|^{2}$, the proposition is proved!