is the problem such easy or am i misunderstanding it? if we have $2|n$ or $3|n$ we can write $n=6+(n-6)$. because of $2|n$ or $3|n$ we have $2|(n-6)$ or $3|(n-6)$ with $n-6\geq 6$ -> finished. if not $2|n$ or $3|n$ we get $n\equiv 1 , 5 \pmod{6}$. if $n\equiv 1$ we set $n=4+(n-4)$, $3|(n-4)$ -> done. if $n\equiv 5$ we set $n=8+n-8$, $3|(n-8)$, $n-8\geq 4$ -> done too. so we have managed all cases and found a solution. Naphthalin

For $ n$ even it is already proved as $ n = 2m ,m\geq6$ $ = (2m - 4) + 4, (2m - 4)\geq8$ which is the sum of two composite numbers. For $ n$ odd we have $ n = 2m + 1 , m\geq6$ $ = (2m - 8) + 9, (2m - 8)\geq4$ which is again the sum of two composite numbers!

Another similar kind of problem is the following: Show that if $ n$ is an even integer $ n\geq40$ then there exists odd composite positive integers $ u$ and $ v$ such that $ n=u+v$.

More than half of the integers smaller than $n$ are composite. So both $k$ and $n-k$ must be composite, for some $k$