Consider number $n = 2^{p-1}$, where $p$ is a prime number. Then $\sigma(n) = 2^p - 1$. Let's prove that $\phi(2^p - 1) > 2^{p-1}$ for infinitely many primes $p$.

Let $m = 2^p - 1 = p_1^{\alpha_1} \dots p_k^{\alpha_k}$. Then
$$\phi(m) = (2^p - 1) \frac{p_1 - 1}{p_1} \dots \frac{p_k - 1}{p_k}.$$
Notice that every $p_i > p$, $i = 1, \dots, k$. Really, as it follows from the little Fermat theorem, $2^{p_i-1} \equiv 1 \pmod{p_i}$, and $2^p - 1 \equiv 0 \pmod{p_i}$ as well, therefore $(p_i - 1) \nmid p$, because $p$ is a prime number.

Hence $m = 2^{p-1} = p_1^{\alpha_1} \dots p_k^{\alpha_k} > p^k$ and so $k \le p \log_p 2$. I.e. $k \le \frac{p}{2}$ for sufficiently large $p$.

Then
$$\frac{\phi(2^p-1)}{2^{p-1}} \ge \frac{2^{p-1}}{2^p} \left(1 - \frac{1}{p}\right)^p.$$
But $\left(1 - \frac{1}{p}\right)^p \to \frac{1}{e}$ as $p \to \infty$, therefore the number $n = 2^{p-1}$ satisfies the condition for sufficiently large $p$.