Example 3-37 A drawer contains 20 shirts, of which 4 are blue, 7 are gray, and 9 are red. How many should be randomly taken to ensure that 4 are of the same color?
How many more should be drawn to ensure that there are 5, 6, 7, 8, 9 shirts of the same color?
According to the Pigeonhole Principle, $n$ pigeonholes, $k n+1$ pigeons, then at least one pigeonhole has $k+1$ pigeons.
(1) Now there are 3 pigeonholes, i.e., $n=3, k+1=4$, so, $k=3$,
$$
k n+1=3 \times 3+1=10
$$

That is, randomly selecting 10 items can ensure that 4 items are of the same color.
(2) It seems that if based on (1), $k=5, n=3$, then one should draw
$$
3 \times 5+1=16
$$

items to ensure 5 items are of the same color. However, this is not the case. The problem model is not entirely the same as the Pigeonhole Principle.
We consider the worst-case scenario: The first 10 items drawn exactly have 4 blue items, i.e., 4 blue items, after which the problem becomes one of finding items of the same color from gray and red. At this point, $n=2, k+1=5, k=4$. Therefore, one should draw $4+$ $4 \times 2+1=13$ items.
(3) $k+1=6, k=5, n=2$,

Thus, one needs to draw
$$
4+(5 \times 2+1)=15 \text { (items) }
$$
(4) $k+1=7, k=6$, one should draw
$$
4+(6 \times 2+1)=17 \text { (items) }
$$
(5) Considering that the blue and gray items have already been taken, only red items remain, $k+1=8, k=7, n=1$, so one should draw
$$
4+7+(k n+1)=4+7+(7+1)=19 \text { (items) }
$$
(6) $k+1==9, k=8, n=1$,
$$
4+7+(k n+1)=4+7+(8+1)=20 \text { (items) }
$$