Among the integers $ n,n+1,n+2,n+3$ at least one is divisible by $ 2$ and at least one is divisible by $ 3$, so these are two of the three prime divisors. The third prime divisor $ p$ is at least $ 5$ so only one of the numbers $ n,n+1,n+2,n+3$ is divisible by $ p$, and the other $ 3$ must be divisible by $ 2$ and/or $ 3$. Among the numbers $ n,n+1,n+2,n+3$ at most two are divisible by $ 3$ so among the three numbers divisible by $ 2$ and/or $ 3$ one is not divisible by $ 3$ and only by $ 2$ so it is a power of $ 2$. By the same reasoning we get one of the numbers $ n,n+1,n+2,n+3$ is a power of $ 3$. Then the number among $ n,n+1,n+2,n+3$ that is not a power of $ 2$ or $ 3$ and not divisible by $ p$ can be either a power of $ 2$, a power of $ 3$, or divisible by $ 2$ and $ 3$. 1. If it is a power of $ 2$ then we get two powers of $ 2$ differing by $ 2$ so they must be $ 2$ and $ 4$. So we either get $ n+3=4$ giving $ n(n+1)(n+2)(n+3)=24$ -- divisible only by $ 2$ and $ 3$ -- so this case fails, or $ n+2=4$ giving $ n=2$ and $ n(n+1)(n+2)(n+3)=120$ -- divisible by $ 2,3,5$ and everything is fine. 2. If it is a power of $ 3$ get two powers of $ 3$ differing by $ 3$ which is impossible. 3. If it is divisible by $ 2$ and $ 3$ the only case when two of the numbers $ n,n+1,n+2,n+3$ are divisible by $ 3$ is when they are $ n$ and $ n+3$. i. $ n$ is a power of $ 3$ ie $ n=3^{k}$, then $ n+3=3^{k}+3$ while $ n+1, n+3$ are even but $ n+3$ is not a power of $ 2$ as it is divisible by $ 3$ so $ n+1$ is a power of $ 2$ ie $ n+1=3^{k}+1=2^{l}$. Also, $ 3^{k}+3=3(3^{k-1}+1)$ and the expression in brackets is not divisible by $ 3$ so it is a power of $ 2$, so get $ 3^{k-1}+1=2^{m}$. Then, $ 3^{k}=(3^{k-1})*3=(2^{m}-1)*3$ but $ 3^{k}=2^{l}-1$ so get $ 3*2^{m}-3=2^{l}-1$ so $ 2^{l-1}+1=3*2^{m-1}$ then $ 2^{m-1}=1$ and $ 2^{l-1}=2$ so $ 2^{l}=4$ so $ 2^{l}-1=3=n$ so $ n=3$. Then $ n(n+1)(n+2)(n+3)=360$ divisible only by $ 2,3,5$ -- everything is fine. ii. $ n+3$ is a power of $ 3$ then $ n+2=2^{l}$, $ n+3=3^{k}$, $ n=3^{k}-3=3(3^{k-1}-1)$ so $ 3^{k-1}-1=2^{m}$ so $ 3*2^{m}+2=2^{l}$ so $ 3*2^{m-1}+1=2^{l-1}$ so $ 2^{m-1}=1$ as it must be odd so $ 2^{l}=4$ so $ n+2=4$ so $ n=2$ and this case works from 1. Therefore, $ n=2$ and $ n=3$ are the only solutions.

n=6，6×7×8×9=(2^4)×(3^3)×7. see: http://www.mathoe.com/dispbbs.asp?boardID=117&ID=16115&page=1

For $n \leqslant 3$, $n = 2,3$ are solutions, now we suppose $n \geqslant 4$. Among the integers $n,n + 1,n + 2,n + 3$， there are two odd numbers $2k - 1,2k + 1,k \geqslant 3$, and $2k - 1,2k,2k + 1$ are pairwise coprimes. So there are exactly three different prime divisors of $2k - 1,2k,2k + 1$,this force $2k$ is power of $2$, and ${2^3}\left| {2k} \right.$. Among the integers $2k - 1,2k,2k + 1$, there is a multiple of $3$, so one of $2k - 1,2k + 1$ must be a power of $3$,hence it is a multiple of $9$. Now we consider the fourth number, it has not any new prime divisor, so it can only divisible by $2,3$. Among the integers $n,n + 1,n + 2,n + 3$， there are not two numbers can divisible by a number great than $3$, so the only possible is $6$, and $n = 6$. Therefore, $n = 2,3,6$ are the only solutions.