Answer: D.

Solution: Let the number of people who participate in all three activities be $x$. We draw a diagram and calculate from the inside out, adding up these parts, we get
$$
39+49+41-[(14-x)+(13-x)+(9-x)]-2 x=100-1 \text {, }
$$

Simplifying, we get
$$
129-(36-3 x)-2 x=99 \text {, }
$$

Solving for $x$, we get
$$
x=6 \text {. }
$$

Therefore, the answer is D.
28. In a Chinese, Math, and English three-subject competition among 100 students at a school, 39 students participated in the Chinese competition, 49 in the Math competition, and 41 in the English competition. 14 students participated in both Chinese and Math competitions, 13 in both Math and English competitions, and 9 in both Chinese and English competitions. 1 student did not participate in any of the three competitions. How many students participated in all three competitions?
(A). 3
(B). 4
(C). 5
(D). 6