Sufficiently Divisible

Let $k$ be the smallest number that satisfies the condition. Since $k+1, k+2, \dots , k+28$ is not divisible by an additional 29, their factorial still satisfies the condition. Until $k+29$ it will be divisible by an additional 29. Hence, the numbers that satisfy the condition is in the range $[k,k+29)$ and there are 29 of them.

Let $n$ be an integer .Total number of $29$ can be found from $n!$ .

$\lfloor \dfrac{n}{29} \rfloor + \lfloor \dfrac{n}{29^2} \rfloor + \lfloor \dfrac{n}{29^3} \rfloor + ......+ \lfloor \dfrac{n}{29^k} \rfloor=871$ where $k$ is the largest factor which can be extracted from $n$ .

Let $n=29^k+m$ for an integer $k,m$ .So, if $m<29$ then

$\lfloor \dfrac{29^k}{29} \rfloor + \lfloor \dfrac{29^k}{29^2} \rfloor + \lfloor \dfrac{29^k}{29^3} \rfloor + ......+ \lfloor \dfrac{29^k}{29^k} \rfloor=871$

$\implies 29^{k-1}+29^{k-2}+29^{k-3}+......+29^2+29^1+1=871 \\ \implies \dfrac{(29^k-1)}{29-1}=871$

Here if $k=3$ satisfies the equation..

If $29 \leq m <57$ then $1$ factor will be extra from the first term.Again $58 \leq m <86$ then $2+1$ factors will be extra.In this way we can vary m and k to get different numbers of factors of $29$ .

Now $n=a=29^3$ is a solution in this case then $a+1,a+2,.....,a+28$ is also the solution because there is no more factor 29 will be there as $29$ is a prime..Again $a+29$ will lead to another one more factor 29 i.e. $872$ factors.So total number of $n$ is $\boxed{29}$