A proof for the even part: For the even numbers, you can say $2\phi(n) \leq n$ easily. Because for each even number the format of $\phi(n)$ is $n(1-\frac{1}{2})c = \frac{n}{2}\times c$ which $c \leq 1$. Hence, $2\phi(n) \leq n$.
For the odd part as $\phi(n) = n(1-\frac{1}{p_1})\cdots (1-\frac{1}{p_k})$, we should consider that such odd numbers which $(1-\frac{1}{p_1})\cdots (1-\frac{1}{p_k}) > \frac{1}{2}$. Hence, all odd numbers which their prime factors satisfy this inequality, they can satisfy the specified inequality. As different set of numbers satisfy this, you can't say any specific about that number unless finding that all said. However, we can say if the prime factor of a number is greater, it has more chance to satisfy the inequality.