An integer \(n\) can be either odd or even. If it is even and greater than 11, then it can be presented as \(4k\) or \(4k+2\), where \(k \ge 3\).
If \(n=4k\), then \(n=2k + 2k\). The only time \(2k\) is prime is when \(k=1\), but we are only considering when \(k \ge 3\), so \(n=4k\) is the sum of composites.
If \(n=4k+2\), then \(n= 2k + 2(k+1)\). Since \(k \ge 3\), then both \(2k\) and \(2(k+1)\) are composite.
What about if \(n\) is odd and greater than 11? Let \(m = n - 9\). Since \(n \gt 11\), then \(m \gt 2\), and since \([\text{odd} - \text{odd} = \text{even}]\), then \(m\) is even. This means \(n-9=2k\), where \(k \gt 2\). Since \(n=2k+9\) and since both 9 and \(2k\) are both composites, then \(n\) is the sum of composites.