I am not an expert here, but I think it has been implicitly considered in the context of $k$-Lehmer numbers.
These are the positive composite integers $n\ge 1$ which satisfy $\phi(n)\mid (n-1)^k$, where
$k\ge 1$ is a fixed positive integer.
There are strong connections between $k$-Lehmer numbers and Carmichael numbers.
Now we have the following elementary observation. We have
$(n-1)^2=(n-1)(n+1)-2(n-1)$. Hence, if $\phi(n)$ divides $2(n-1)$ and
$n^2-1$, then $n$ is a $2$-Lehmer number.
It is easy to see that there always exist $k$-Lehmer numnbers for all $k\ge 2$. Of course, Lehmer's conjecture is, that there are no $1$-Lehmer numbers.
I know, this does not imply too much with respect to your question, but I tend to believe, that there
should be no composite solutions to your equation $\phi(n)\mid 2(n-1)$, too. As far as I know, this is not known.
For some more interesting facts about $k$-Lehmer numbers, see, for example, the article http://arxiv.org/abs/1012.2337.