[b]a.)[/b] For which $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers?

[b]b.)[/b] For which $n>2$ is there exactly one set having this property?
### Part (a)

1. **Show that there is no such set for \( n = 3 \):**

   Suppose there exists a set of 3 consecutive positive integers \(\{a, a+1, a+2\}\) such that the largest number \(a+2\) is a divisor of the least common multiple (LCM) of the remaining two numbers \(\{a, a+1\}\).

   - The LCM of \(\{a, a+1\}\) is \(\text{LCM}(a, a+1)\).
   - Since \(a\) and \(a+1\) are consecutive integers, they are coprime, so \(\text{LCM}(a, a+1) = a(a+1)\).

   For \(a+2\) to be a divisor of \(a(a+1)\), \(a+2\) must divide \(a(a+1)\). However, for any integer \(a\), \(a+2\) is not a divisor of \(a(a+1)\) because \(a+2\) is not a factor of either \(a\) or \(a+1\). This leads to a contradiction.

   Therefore, there is no such set for \( n = 3 \).

2. **Show that there always exists a set for \( n \geq 4 \):**

   Consider the set of \( n \) consecutive integers \(\{a, a+1, \ldots, a+n-1\}\). Let \( k_n = a+n-1 \) be the largest element in the set.

   - We need \( k_n \) to be a divisor of the LCM of the remaining \( n-1 \) numbers.
   - For \( n \geq 4 \), we can choose \( a \) such that \( k_n \) is a product of prime powers that are less than \( n \).

   For example, for \( n = 4 \), consider the set \(\{3, 4, 5, 6\}\):
   - The LCM of \(\{3, 4, 5\}\) is \( \text{LCM}(3, 4, 5) = 60 \).
   - The largest number \( 6 \) divides \( 60 \).

   This logic can be extended to any \( n \geq 4 \) by choosing appropriate sets of consecutive integers.

   Therefore, there always exists a set for \( n \geq 4 \).

### Part (b)

1. **Show that \(\{3, 4, 5, 6\}\) is the unique set for \( n = 4 \):**

   Consider the set \(\{a, a+1, a+2, a+3\}\) where \( k_4 = a+3 \).

   - For \( k_4 \) to be a divisor of the LCM of the remaining three numbers, \( k_4 \) must be a product of prime powers that are less than 4.
   - The only primes less than 4 are 2 and 3.

   The highest possible power of 2 is 1, and the highest possible power of 3 is also 1. Therefore, \( k_4 \) must be \( 2 \times 3 = 6 \).

   The only set of 4 consecutive integers ending in 6 is \(\{3, 4, 5, 6\}\).

   Therefore, \(\{3, 4, 5, 6\}\) is the unique set for \( n = 4 \).

2. **Show that there exist more than one set for \( n > 4 \):**

   For \( n > 4 \), consider the set \(\{a, a+1, \ldots, a+n-1\}\) where \( k_n = a+n-1 \).

   - Let \( p_x \) be the largest prime less than \( n \).
   - Set all \( a_i \) to 1, so \( k_n = \prod_{i=1}^x p_i \).

   This works because \( k_n = \prod_{i=1}^x p_i > n \). Additionally, we can set \( a_1 = 2 \), which also works.

   Therefore, there exist more than one set for \( n > 4 \).

The final answer is \( \boxed{ n \geq 4 } \) for part (a) and \( n = 4 \) for part (b).