Let real numbers $a_{1}, a_{2}, \cdots, a_{2013}$ (allowing duplicates) have an arithmetic mean of $m$. A set $\{i, j, k\}$ is called a good set if it satisfies
$$
a_{i}+a_{j}+a_{k} \geqslant 3 m \quad(1 \leqslant i<j<k \leqslant 2013)
$$

Find the minimum possible number of good sets.
(Lin Chang provided the problem)
Consider $n=3 k\left(k \in \mathbf{N}^{*}\right), \sum_{i=1}^{n} a_{i}=3 k m$.

Divide $a_{1}, a_{2}, \cdots, a_{n}$ into $k$ triples arbitrarily. Since the sum of the $k$ sums equals $3 k m$, at least one of them is no less than $3 m$, i.e., there is at least one good group.
The number of ways to divide $3 k$ numbers into triples is $\frac{(3 k)!}{(3!)^{k} \cdot k!}$. Each division yields at least one good group. On the other hand, each good group appears as many times as the number of ways to divide the remaining $3 k-3$ numbers into triples, which is $\frac{(3 k-3)!}{(3!)^{k-1} \cdot(k-1)!}$. Therefore, the number of good groups is no less than
$$
\frac{(3 k)!}{(3!)^{k} \cdot k!} \div \frac{(3 k-3)!}{(3!)^{k-1} \cdot(k-1)!}=\frac{(3 k-1)(3 k-2)}{2}=\mathrm{C}_{n-1}^{2} .
$$

Take $a_{1}=a_{2}=\cdots=a_{n-1}=m-1, a_{n}=m+n-1$, then any triple is a good group if and only if it contains $a_{n}$, so the number of good groups is $\mathrm{C}_{n-1}^{2}$, achieving the minimum value.
In particular, for $n=2013$, the result is $\mathrm{C}_{2012}^{2}$.