14.3.16 ** (1) For what integers $n>2$, are there $n$ consecutive positive integers, where the largest number is a divisor of the least common multiple of the other $n-1$ numbers?
(2) For what integers $n>2$, is there exactly one set of positive integers with the above property?
Let $n$ consecutive positive integers have the largest one as $m$. When $n=3$, if $m$ is a divisor of the least common multiple of $m-1, m-2$, then $m \mid (m-1)(m-2)$, which implies $m \mid 2$, which is impossible. When $n=4$, from $m \mid (m-1)(m-2)(m-3)$, we know $m \mid 6$, and since $m>4$, the only set of positive integers with the given property is $3,4,5,6$. Suppose $n>4$, take $m=(n-1)(n-2)$, then $n-1 \mid m-(n-1), n-2 \mid m-(n-2)$. Since $n-1$ and $n-2$ are coprime, $m=(n-1)(n-2)$ can divide the least common multiple of $m-(n-1)$ and $m-(n-2)$, so $m$ has the given property. Similarly, taking $m=(n-2)(n-3)$, $m$ can divide the least common multiple of $m-(n-2)$ and $m-(n-3)$, so $m$ has the given property. Therefore, when $n \geqslant 4$, there always exists a set of positive integers with the given property. Only when $n=4$ is there exactly one such set of positive integers.