8. (12 points) In a cage, there are 21 crickets and 30 grasshoppers. The Red-Haired Magician, with each transformation, turns 2 grasshoppers into 1 cricket; the Green-Haired Magician, with each transformation, turns 5 crickets into 2 grasshoppers. After the two magicians have transformed a total of 15 times, there are only grasshoppers left in the cage, with no crickets. At this point, there are $\qquad$ grasshoppers.
【Solution】Solution: According to the problem, we can get: every time the Red-Haired Magician performs a transformation, 1 more cricket appears; every time the Green-Haired Magician performs a transformation, 5 fewer crickets appear;

Since the two magicians together performed 15 transformations after which only katydids remained in the cage with no crickets, the original number of crickets in the cage plus the number of crickets added by the Red-Haired Magician after several transformations must be a multiple of 5,

Therefore, the number of times the Red-Haired Magician performs the transformation is $4, 9$ or 14;
(1) When the Red-Haired Magician performs the transformation 4 times,
The number of times the Green-Haired Magician performs the transformation is: $15-4=11$ (times);
$$
\begin{aligned}
& (21+4) \div 5 \\
= & 25 \div 5 \\
= & 5 \text { (times) }
\end{aligned}
$$

Since $5 \neq 11$,
Therefore, when the Red-Haired Magician performs the transformation 4 times, it does not meet the problem's requirements.
(2) When the Red-Haired Magician performs the transformation 9 times,
The number of times the Green-Haired Magician performs the transformation is: $15-9=6$ (times);
$$
\begin{aligned}
& (21+9) \div 5 \\
= & 30 \div 5 \\
= & 6 \text { (times) }
\end{aligned}
$$

Therefore, when the Red-Haired Magician performs the transformation 9 times, it meets the problem's requirements, at this time the number of katydids is:
$$
\begin{array}{l}
30-2 \times 9+2 \times 6 \\
=30-18+12 \\
=24 \text { (individuals) }
\end{array}
$$
(3) When the Red-Haired Magician performs the transformation 14 times,
The number of times the Green-Haired Magician performs the transformation is: $15-14=1$ (time);
$$
\begin{aligned}
& (21+14) \div 5 \\
= & 35 \div 5 \\
= & 7 \text { (times) }
\end{aligned}
$$

Since $1 \neq 7$,
Therefore, when the Red-Haired Magician performs the transformation 14 times, it does not meet the problem's requirements.
In summary, we can get
When the Red-Haired Magician performs the transformation 9 times, and the Green-Haired Magician performs the transformation 6 times, only katydids remain in the cage with no crickets, at this time there are 24 katydids.

Answer: At this time, there are 24 katydids.
The answer is: 24.