Answer: none.

Solution. Let for some natural numbers $m$ and $n$ the equality $n(n+1)(n+2)(n+3)+m^{5}=5000$ holds. Since $n(n+1)(n+2)(n+3)=\left(n^{2}+3 n+1\right)^{2}-1$,
the equality can be rewritten as $k^{2}+m^{5}=5001$, where $k=n^{2}+3 n+1$. Consider the remainders upon division by 11 of the left and right parts of the equality $k^{2}+m^{5}=5001$. Squares can only give remainders of $0,1,3,4,5$ and 9, while fifth powers can only give remainders of 0,1, and 10. But no two of the written remainders add up to the remainder 6, which the number 5001 will have.