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$\Large\frac{n^5 - 5n^3 + 4n}{n+2}=\frac{(n-2)(n-1)(n)(n+1)\color{#D61F06}{(n+2)}}{\color{#D61F06}{(n+2)}}$ $\Large =(n-2)(n-1)(n)(n+1)$ which has lowest positive integral value as $\large\boxed{24}$ which must also be the greatest integer which divides the given expression for all natural numbers n.(Induction can be used to prove that (n-2)(n-1)(n)(n+1) is divisible by 4 for all natural numbers 'n'.. Refer here for a proof)

We note first that

$n^{5} - 5n^{3} + 4n = n(n^{4} - 5n^{2} + 4) = n(n^{2} - 4)(n^{2} - 1) = n(n - 2)(n + 2)(n - 1)(n + 1)$ ,

which when divided though by $n + 2$ and then rearranged yields $(n - 2)(n - 1)(n)(n + 1)$ .

For both $n = 1$ and $n = 2$ the given expression equals $0$ , so we can now focus on looking for the greatest integer $m$ that divides the product of any $4$ consecutive positive integers. Now, at the very least, for any such product one of the terms will be divisible by $2$ and another by $4$ , and at least one will be divisible by $3$ . Thus $m$ must divide $2*4*3 = 24$ , and since for $n = 3$ the given expression comes out to $24$ we can have $m$ no greater than $24$ . Thus $m = \boxed{24}$ .