polynomial with positive integer coefficients divisible by 24?

$n(n+1)(n+2)(n+3)$ is divisible by $8$ because there are two even numbers and one of them is divisible by $4$.

$$n^4+6n^3+11n^2+6n=n(n^3+6n^2+11n+6)=n(n+1)(n^2+5n+6)$$

$$=n(n+1)(n+2)(n+3)$$ which is a Product of $4$ consecutive integers

Now see The product of n consecutive integers is divisible by n factorial

We have $$ n^4+6n^3+11n^2+6n=n(n+1)(n+2)(n+3)=24\binom{n+3}4. $$ This settles the claim when $n\ge1$. To get it for all $n$ observe that all integers are congruent to one $>1$ modulo $24$.