8. From $1,2, \cdots, 2013$, select $3 k$ different numbers to form $k$ triples $\left(a_{i}, b_{i}, c_{i}\right)(i=1,2$, $\cdots, k)$. If $a_{i}+b_{i}+c_{i}(i=1,2, \cdots, k)$ these $k$ numbers are all distinct and less than 2013, then the maximum value of $k$ is
8. 402.

On the one hand,
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \leqslant \sum_{i=1}^{k}(2013-i) \\
=2013 k-\frac{k(k+1)}{2},
\end{array}
$$

and
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \geqslant 1+2+\cdots+3 k \\
=\frac{3 k(3 k+1)}{2} .
\end{array}
$$

Thus, $2013 k-\frac{k(k+1)}{2} \geqslant \frac{3 k(3 k+1)}{2}$.
Solving this, we get $k \leqslant 402$.
On the other hand, construct
$$
\left(a_{i}, b_{i}, c_{i}\right)=(i, 402+i, 1207-i),
$$

where, $(i=1,2, \cdots, 402)$.
The sum of the three numbers in this triplet is $1609+i$ $(i=1,2, \cdots, 402)$, which satisfies the problem's conditions.