3. Several tourists are to board cars, with the requirement that the number of people in each car is equal. If each car carries 28 people, then 1 person is left without a car; if one car is reduced, then all tourists can be evenly distributed among the remaining cars. Given that each car can accommodate a maximum of 35 people, then there are $\qquad$ tourists and $\qquad$ cars.
$3.841,30$.
Suppose there are $k$ cars, and later each car carries $x$ people, then we have $28 k+1=n(k-1)$.
Thus, $n=\frac{28 k+1}{k-1}=28+\frac{29}{k-1}$.
Therefore, $k-1=1$ or $k-1=29$, which means $k=2$ or $k=30$. When $k=2$, $n=57$ is not reasonable, so we discard it.
Upon verification, we find that $k=30, n=29$.
Therefore, there are 30 cars, and the number of tourists is $29 \times 29=841$ (people).