2・ 10 (1) For what natural numbers $n>2$, is there a set of $n$ consecutive natural numbers such that the largest number in the set is a divisor of the least common multiple of the other $n-1$ numbers?
(2) For what $n>2$, is there exactly one set with the above property?
[Solution] (1) First, we point out that $n \neq 3$. Otherwise, let $\{r, r+1, r+2\}$ have the property required by the problem: $(r+2) \mid r(r+1)$. Since $(r+1, r+2)=1$, it follows that $(r+2) \mid r$, which is a contradiction.

Below, we discuss the cases for $n \geqslant 4$ by considering whether $n$ is odd or even.
(i) If $n=2k$, since $n-1$ is odd, the set of $n$ consecutive natural numbers $\{n-1, n, n+1, \cdots, 2(n-1)\}$ satisfies the requirement.
(ii) If $n=2k+1$, then $n-2$ is odd, so the set of $n$ consecutive natural numbers $\{n-3, n-2, n-1, \cdots, 2(n-2)\}$ satisfies the requirement.

From (i) and (ii), we can see that when $n \geqslant 4$, there is always a set of $n$ consecutive natural numbers that satisfies the requirement.
(2) We first prove that when $n \geqslant 5$, there are at least two sets of $n$ consecutive natural numbers that satisfy the requirement. $\square$
(i) When $n$ is an even number not less than 6, in addition to the set given in (i), the set $\{n-5, n-4, \cdots, 2(n-3)\}$ also satisfies the requirement.
(ii) When $n \geqslant 9$ is an odd number, in addition to the set given in (1), the set $\{n-7, n-6, \cdots, 2(n-4)\}$ also satisfies the requirement.
(iii) When $n=5$, both $\{2,3,4,5,6\}$ and $\{8,9,10,11,12\}$ satisfy the requirement; when $n=7$, both $\{4,5,6,7,8,9,10\}$ and $\{6,7,8,9,10,11,12\}$ satisfy the requirement.

Finally, we prove that when $n=4$, the only set $\{3,4,5,6\}$ satisfies the requirement.

Suppose $\{k, k+1, k+2, k+3\}$ satisfies the requirement, i.e., $k+3$ divides the least common multiple of $k, k+1, k+2$. Since $(k+2, k+3)=1$, it follows that $k+3$ divides the least common multiple of $k, k+1$. Since $(k, k+1)=1$, it follows that $(k+3) \mid k(k+1)$.

If $k+3$ is odd, then $k+1$ and $k+3$ are two consecutive odd numbers, so $(k+1, k+3)=1$. Therefore, $(k+3) \mid k(k+1)$ implies $(k+3) \mid k$, which is a contradiction.

If $k+3$ is even, let $k+3=2m$, then we have
$2m \mid 2(m-1)(2m-3)$.
This implies $m \mid 2m-3$, hence $m \mid 3$. Since $m \neq 1$, it must be that $m=3$, which proves that $\{3,4,5,6\}$ is the only set that satisfies the requirement.