$f(x)=\pi(n+999)-\pi(n)$ is continuous in the following meaning: if $n$ changes by $1$, $f(n)$ changes by at most $1$. If it changes from $a$ to $b$, it did have every value $c$ between them at least once (everything integers). With $a=f(2) > 5$, $b=f(1001!+1) =0$, we can find $n$ such that there are exactly $5$ primes in $\{n,n+1,...,n+999\}$.

I apologize to everyone for my solution being very similar to that of ZetaX. Let $f(n)(n\in \mathbb N)$ denotes the number of prime numbers from $\{ n,\dots ,n+999\}$.It is easy to show that $f(n+1)-f(n)=0,\pm 1$.Obviously $f(1)>5$ and $f(1001!+2)=0$.Then by intermediate value theorem,$1<$∃$m<1001!+2$ s.t. $f(m)=5$.Therefore there exists $1000$ consecutive positive integers containing exactly five prime numbers.$\blacksquare$