3. When $n=2 m$, take
$$
\begin{array}{l}
z_{1}=z_{2}=\cdots=z_{m}=1, \\
z_{m+1}=z_{m+2}=\cdots=z_{2 m}=-1 ;
\end{array}
$$

When $n=2 m+1$, take
$$
\begin{array}{l}
z_{1}=z_{2}=\cdots=z_{m}=1, \\
z_{m+1}=z_{m+2}=\cdots=z_{2 m+1}=-\frac{m}{m+1} .
\end{array}
$$

In both cases, we have $c \geqslant \frac{2}{n}$.
We now prove that for any complex numbers $z_{1}, z_{2}, \cdots, z_{n}$, we have
$$
\left|\sum_{i=1}^{n} z_{i}\right|+\frac{2}{n} \sum_{1 \leqslant i<j \leqslant n}\left|z_{i}-z_{j}\right| \geqslant \sum_{i=1}^{n}\left|z_{i}\right| .
$$

In fact, for $1 \leqslant k \leqslant n$, we have
$$
\begin{array}{l}
\left|\sum_{i=1}^{n} z_{i}\right|+\sum_{j=1}^{n}\left|z_{k}-z_{j}\right| \\
\geqslant\left|\sum_{i=1}^{n} z_{i}+\sum_{j=1}^{n}\left(z_{k}-z_{j}\right)\right|=n\left|z_{k}\right| .
\end{array}
$$

Summing over $k$ from 1 to $n$ yields
$$
n\left|\sum_{i=1}^{n} z_{i}\right|+2 \sum_{1 \leqslant k<j \leqslant n}\left|z_{k}-z_{j}\right| \geqslant n \sum_{k=1}^{n}\left|z_{k}\right| \text {. }
$$

In conclusion, the smallest positive real number is $\frac{2}{n}$.
3. Given an integer $n \geqslant 2$. Find the smallest positive real number $c$, such that for any complex numbers $z_{1}, z_{2}, \cdots, z_{n}$, we have
$$
\left|\sum_{i=1}^{n} z_{i}\right|+c \sum_{1 \leqslant i<j \leqslant n}\left|z_{i}-z_{j}\right| \geqslant \sum_{i=1}^{n}\left|z_{i}\right| .
$$