24 From $1,2,3, \cdots, 999,1000$, arbitrarily select $k$ numbers. If among the selected numbers, there are always three numbers that can form the side lengths of a triangle, what is the minimum value of $k$? And explain the reason.
The minimum value of $24 k$ is 16.
Assume the minimum value of $k$ is $>16$, then among the 16 selected numbers $a_{1}<a_{2}<\cdots<a_{16}$, we have
$$
\begin{array}{c}
a_{1} \geqslant 1, \quad a_{2} \geqslant 2, \\
a_{3} \geqslant a_{1}+a_{2} \geqslant 3, \\
a_{4} \geqslant a_{2}+a_{3} \geqslant 5, \\
\cdots \cdots \\
a_{15} \geqslant a_{13}+a_{14} \geqslant 987, \\
a_{16} \geqslant a_{14}+a_{15} \geqslant 1597 .
\end{array}
$$

This contradicts $a_{16} \leqslant 1000$. Therefore, $k$ can take 16.
Furthermore, from $\{1,2, \cdots, 1000\}$, selecting 15 numbers $\{1,2,3,5,8,13,21,34,55,89,144,233,377,610,987\}$.
No three numbers among these 15 can form a triangle, so $k \neq 15$. That is, $k \geqslant 16$.
In conclusion, the minimum value of $k$ is 16.