Three, (25 points) 120 people participated in a math competition, with the test consisting of 5 major questions. It is known that questions $1, 2, 3, 4, 5$ were correctly solved by 96, $83, 74, 66, 35$ people respectively. If at least 3 questions must be answered correctly to win an award, how many people won awards at least in this competition?
Three, number these 120 people as $P_{1}, P_{2}, \cdots, P_{120}$, and consider them as 120 points on a number line. Let $A_{k}(k=1, 2,3,4,5)$ represent the group of people among these 120 who did not answer the $k$-th question correctly, and $\left|A_{k}\right|$ be the number of people in that group. Then,
$\left|A_{1}\right|=24,\left|A_{2}\right|=37,\left|A_{3}\right|=46$,
$\left|A_{4}\right|=54,\left|A_{5}\right|=85$.
Assign five different colors to the five groups. If a person did not answer the $k(k=1,2,3,4,5)$-th question correctly, then the point representing that person is colored with the $k$-th color.

The problem is transformed into: finding the maximum number of points that are colored with at least three colors.

From $\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{4}\right|+\left|A_{5}\right|=246$, we know that the number of points colored with at least three colors is no more than $\frac{246}{3}=82$.

On one hand, color the points $P_{1}, P_{2}, \cdots, P_{85}$ with the 5th color. Since $85>82$, to maximize the number of points colored with at least three colors, we need to color as many of these 85 points with two additional colors as possible. Given that
$$
\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{4}\right|=161 \text {, }
$$
the number of points that can be colored with two additional colors will not exceed $\left[\frac{161}{2}\right]=80$, meaning the number of points colored with at least three colors will not exceed 80.

On the other hand, we can construct a specific coloring method to ensure that 80 points are colored with three colors. For example, as shown in Figure 6, color the points $P_{1}, P_{2}, \cdots, P_{85}$ with the 5th color; color the points $P_{1}$, $P_{2}, \cdots, P_{44}$ and $P_{79}, P_{80}$ with the 3rd color; color the points $P_{45}, P_{46}, \cdots, P_{81}$ with the 2nd color; color the points $P_{1}, P_{2}, \cdots, P_{24}$ with the 1st color; and color the points $P_{25}, P_{26}, \cdots, P_{78}$ with the 4th color. Thus, the maximum number of points colored with at least three colors is 80. Therefore, the number of points colored with no more than two colors is at least 40, meaning at least 40 people win (each of them answered at most two questions incorrectly and at least three questions correctly, for example, $P_{81}, P_{82}, \cdots, P_{120}$ these 40 people).