2. Given an integer $n \geqslant 2$, find the largest constant $C_{n}$, such that for any real numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}+\cdots+a_{n}=0$, we have $\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\} \leqslant C_{n} \cdot \min _{1 \leqslant i \leqslant n} a_{i}$ (here we assume $a_{n+1}=a_{1}$ ).
Take $a_{1}=1-n, a_{2}=a_{3}=\cdots=a_{n}=1$, then we have
$\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\}=2(1-n)+(n-2)=-n \leqslant C_{n} \cdot \min _{1 \leqslant i \leqslant n} a_{i}=C_{n} \cdot(1-n)$
$\Rightarrow n \geqslant C_{n}(n-1) \Rightarrow C_{n}=\frac{n}{n-1}$. We now prove that the constant $C_{n}=\frac{n}{n-1}$ satisfies the condition.
Considering that for a given set of real numbers $a_{1}, a_{2}, \cdots, a_{n}, C_{n} \cdot \min _{1 \leqslant i \leqslant n} a_{i}$ has a fixed value, it is only necessary to ensure that the maximum value of $\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\}$ meets the condition. Since there are only a finite number of different circular arrangements of $a_{1}, a_{2}, \cdots, a_{n}$, the maximum value of $\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\}$ must exist.
Suppose $a_{1}, a_{2}, \cdots, a_{n}$ have $k(1 \leqslant k \leqslant n)$ different values $b_{1}, b_{2}, \cdots, b_{k}$, and $b_{1}<b_{2}<\cdots<b_{k}$, where $b_{1}$ appears $\lambda_{1}$ times, $b_{2}$ appears $\lambda_{2}$ times, $\cdots, b_{k}$ appears $\lambda_{k}$ times, then $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}=n$.
In the sum $\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\}$, $b_{k}$ can appear at most $\lambda_{k}-1$ times, $b_{k-1}, b_{k}$ together can appear at most $\lambda_{k-1}+\lambda_{k}-1$ times, similarly, $b_{k-2}, b_{k-1}, b_{k}$ together can appear at most $\lambda_{k-2}+\lambda_{k-1}+\lambda_{k}-1$ times, $\cdots, b_{2}, b_{3}, \cdots, b_{k-1}, b_{k}$ together can appear at most $\lambda_{2}+\lambda_{3}+\cdots+\lambda_{k-1}+\lambda_{k}-1$ times, and $b_{1}, b_{2}, \cdots, b_{k-1}, b_{k}$ together appear exactly $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k-1}+\lambda_{k}=n$ times.
Therefore, to maximize $\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\}$, $b_{k}$ must appear exactly $\lambda_{k}-1$ times, $b_{k-1}$ must appear exactly $\lambda_{k-1}$ times, $\cdots, b_{2}$ must appear exactly $\lambda_{2}$ times, and $b_{1}$ must appear exactly $\lambda_{1}+1$ times. This can be achieved by arranging $a_{1}, a_{2}, \cdots, a_{n}$ in descending order clockwise on a circle (equal numbers adjacent).
At this point, $\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\}=\sum_{i=1}^{k} \lambda_{i} b_{i}+b_{1}-b_{k}=\sum_{i=1}^{n} a_{i}+b_{1}-b_{k}=b_{1}-b_{k}$ is the maximum value.
Since $b_{1}-b_{k} \leqslant \frac{n}{n-1} \cdot b_{1} \Leftrightarrow b_{k}-b_{1} \geqslant-\frac{n b_{1}}{n-1}$
$$
\Leftrightarrow(n-1) b_{k}+b_{1} \geqslant\left(\sum_{i=1}^{k} \lambda_{i} b_{i}-b_{1}\right)+b_{1}=\sum_{i=1}^{n} a_{i}=0 \text {, }
$$

Thus, $\sum_{i=1}^{n} \min \left\{a_{i}, a_{i+1}\right\} \leqslant \frac{n}{n-1} \cdot \min _{1 \leqslant i \leqslant n} a_{i}$, so the largest constant $C_{n}=\frac{n}{n-1}$.