8、We know that the number of divisors of $2013$, $2014$, and $2015$ are the same. Therefore, for three consecutive natural numbers $n$, $n+1$, and $n+2$ that have the same property (the same number of divisors), the smallest value of $\mathbf{n}$ is
【Answer】 33
Analysis: Three consecutive numbers cannot all be prime numbers. To make their number of factors the same, the following must be achieved:
(1) None of them are prime numbers (otherwise the number cannot be equal);
(2) There cannot be a perfect square among the three numbers (otherwise the number of factors would be odd and even, and cannot be equal).

For extremum problems, start from the extreme situation, from small to large, and eliminate prime numbers and perfect squares, as shown below:
$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,$
$23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 \ldots \ldots$
After testing, $33, 34, 35$ each have 4 divisors, $\mathrm{n}$ is the smallest at 33