When is the Fourier transform of a signal periodic?
I'm new to this exchange and I'm not sure how mathy you all get. I think the answer below is cool because it shows that in some sense the continuous-time Fourier transform is never periodic but that in another sense there are lots of ways to get periodic transforms. 

For the continuous-time Fourier transform on $\mathbb{R}$, both CMDoolittle's and Robert Bristow-Johnson's answers are correct, though for slightly-different definitions of "Fourier Transform". In the usual way we talk about the continuous Fourier transform, transforms are never periodic. If we're willing to transform distributions (of which the Dirac delta is an example), we can have transforms which are periodic.

Usually, the theory of the continuous Fourier transform is developed on $L^2(\mathbb{R})$. The Fourier transform is a bijection of $L^2(\mathbb{R})$ back onto itself; this means that $L^2(\mathbb{R})$ is also the space of all possible Fourier transforms. However, the zero function is the only periodic function in $L^2(\mathbb{R})$, so we can conclude that continuous Fourier transforms of non-zero functions are never periodic. This is CMDoolittle's answer.

However, we also have a Fourier theory for distributions. Specifically, we define distributions to be the set of linear functionals on the space of so-called test functions. Test functions, in turn, are $C^\infty$ functions which are required to decay rapidly (alternately, sometimes test functions are $C^\infty$ functions with compact support). Some of these distributions can be identified with periodic functions on $\mathbb{R}$. For example, if $f$ is a test function then the mapping 
$$ S : f \mapsto \int_{-\infty}^\infty \sin(t) f(t)\ dt$$
is a distribution which can be identified with the sine function. As we can define the Fourier transform of a distribution, so we have a way of defining the Fourier transform of periodic functions. As it turns out, the Fourier transform is also a bijection of the space of distributions back onto itself. Going back to the distribution $S$, this means that we can write $S$ as the Fourier transform of some other distribution $T$: $S = \hat{T}$; in other words, we conclude that there has to be a distribution $T$ whose transform is the distribution $S$ which is identified with the periodic function sine. Clearly, we could pull this same trick with, say, any bounded periodic function. Ultimately then, we can extend Robert Bristow-Johnson's answer: if we're talking about the Fourier transform of distributions, there are lots of distributions whose Fourier tranforms are periodic.

For a discussion of the Fourier theory of distributions, see Chapter 7 of Rudin's Functional Analysis.