Given a positive integer $n$, there are $3 n$ numbers satisfying:
$$
\begin{array}{c}
0 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{3 n}, \\
\text { and }\left(\sum_{i=1}^{3 n} a_{i}\right)^{3} \geqslant k_{n}\left(\sum_{i=1}^{n} a_{i} a_{n+i} a_{2 n+i}\right) \text { always holds. }
\end{array}
$$

Find the best possible value of $k_{n}$ (expressed in terms of $n$).
Lemma 1 Let $0 \leqslant x \leqslant K \leqslant y$. Then $x y \leqslant K(x+y-K)$.
Proof of Lemma 1 Clearly,
$$
(K-x)(y-K) \geqslant 0 \text {. }
$$

Lemma 2 Let $0 \leqslant x \leqslant K \leqslant y \leqslant L \leqslant z$. Then $x y z \leqslant K L(x+y+z-K-L)$.
Proof of Lemma 2 In fact,
$$
\begin{array}{l}
x y z \leqslant K(x+y-K) z \\
\leqslant K L((x+y-K)+z-L) \\
=K L(x+y+z-K-L) .
\end{array}
$$

Lemmas 1 and 2 are proved.
Let $S=\sum_{i=1}^{3 n} a_{i}, K=a_{n}, L=a_{2 n}$.
Then by Lemma 1 we get
$$
\begin{array}{l}
\sum_{i=1}^{n} a_{i} a_{n+i} a_{2 n+i} \\
\leqslant \sum_{i=1}^{n} K L\left(a_{i}+a_{n+i}+a_{2 n+i}-K-L\right) \\
=K L(S-n K-n L) .
\end{array}
$$

Using the AM-GM inequality, we get
$$
\begin{array}{l}
n K \cdot n L(S-n K-n L) \\
\leqslant\left(\frac{n L+n K+(S-n K-n L)}{3}\right)^{3}=\frac{S^{3}}{27} .
\end{array}
$$

When $a_{i}=1$, the equality holds in all the above inequalities. Therefore, the best value of $k_{n}$ is $27 n^{2}$.