Why is conformal invariance only possible for massless theories? I'm conscious that this isn't necessarily a research level question, but I've asked this question on mathstackexchange, and received no answer. So I'm trying it here.
A usual mantra in field theories is the assertion that only massless theories can be conformally invariant. By a theory I mean an action $$ S = \int \mathcal{L} \, \mathrm{dVol}, $$ where $\mathcal{L}$ is the Lagrangian density, and the integral is taken over a 4-dimensional Lorentzian manifold with metric $g$. By conformal invariance I mean the statement that under the conformal rescaling of the metric $$ \hat{g} = \Omega^2g, $$ the Lagrangian transforms as $\hat{\mathcal{L}} = \Omega^{-4} \mathcal{L}$. Then, as the volume form transforms as $\widehat{\mathrm{dVol}} = \Omega^{4} \mathrm{dVol}$, the action $S$ is invariant, and the theory is said to be conformally invariant.
The usual physics explanation given is that "if a theory is supposed to be conformally invariant, then there cannot exist an intrinsic scale to it, such as mass or a Compton wavelength". Of course, this is a load of hand waving. I guess I don't strictly know what I mean by a massless theory. Maxwell's equations, for example, are a massless conformally invariant theory. My guess would have been that the mass of a theory is its ADM mass, but as has been pointed out in the comments, this is a property of a solution to a theory, not the theory itself. So, if $m$ is the mass of a theory, whatever it stands for exactly, and $m \neq0$, why must conformal invariance fail?
 A: Note that the Schrödinger action, say in flat space-time,
$$
S=\int\mathrm dx\ \psi^*\left(i\frac{\mathrm d}{\mathrm dt}+\frac{1}{2m}\nabla^2\right)\psi
$$
has a length scale (mass) $m$, and yet it is scale invariant. In fact, it is invariant under the so-called Schrödinger group, which includes the centrally extended Galilei Group (the Bargmann Group) together with dilatations and conformal transformations.
Therefore, absence of dimension-full constants is not a requisite for conformal and/or dilatation invariance.
A: Why is "conformal invariance only possible for massless theories?" 
Actually, the answer is NO. 
Not only the massless, but also the infinite large massive theory can have "conformal invariance."  As long as the theory sets no scale in order to be scale invariance.
Some massless theories
(or gapless, which means the energy gap $\Delta_E$ for excitations away from the vacuum ground state is zero, $$\Delta_E\equiv E_{exc}- E_0 \to 0,$$ where $E_{exc}$ and $E_0$ are the energies of excited states and ground state respectively) theory can be "conformal invariance" or be a "conformal field theory (CFT)." The $\Delta_E=0$ sets no scale in the theory, thus it can be scale invariant.
The infinite large massive theory, or the infinite large gap theory can still be "conformal invariance" or be a "conformal field theory." The infinite large massive theory means there is an infinitely large mass $M\to \infty$, or say the energy gap $\Delta_E$  for excitations away from the vacuum ground state has 
$$\Delta_E\equiv E_{exc}- E_0 \to \infty.$$
So the length/time scale of the system (based on the dimensional analysis $\Delta_p \Delta_x \sim \Delta_E \Delta_t \sim \hbar$) $$\Delta_l \to 0$$
The well-known examples of such a massive $\Delta_E \to \infty$ is Topological Quantum Field Theory (TQFT). Indeed, Topological Quantum Field Theory at its ground state has no scales (thus scale invariance) and it is also Poincaré (translations + Lorentz) invariance, and also have a special conformal transformation (if needed). Thus TQFT at ground state is also conformal invariant. The $\Delta_E=\infty$ also sets no scale in the theory, thus it can be scale invariant.
TQFTs may be among the most simplest/trivial examples of CFT. But TQFTs are still complicated enough to have many different distinct classes.
The "conformal invariance" requires the scale invariance/dilations and the special conformal transformation, and Poincaré group (the semi direct product of translations group and Lorentz group).
A: A physicist would answer this question as follows.  (Everything I'll say can be expressed in a way that the purest of mathematicians would understand, but that translation would take a lot of work, so I'll only do it on demand.)
In physics we have units of mass ($M$), length ($L$) and time ($T$).  
In special relativity we have a fundamental constant $c$, the speed of light, with units $L/T$.  Thus, in special relativity, any length determines a time and vice versa.  
In quantum mechanics we have a fundamental constant $\hbar$, Planck's constant, with units $ML^2/T$.   $\hbar / c$ has units $M L$.  Thus, in theories involving both special relativity and quantum mechanics, any mass determines an inverse length, and vice versa.
Relativistic quantum field theory involves both special relativity and quantum mechanics.  Thus, in relativistic quantum field theory, if we have a particle of some mass $m$, it determines a length, namely $\hbar / c m$.  This is called the 
Compton wavelength of that particle.  
Some physicists may prefer to use $h = 2 \pi \hbar$ in the definition of the Compton wavelength, but this isn't important here: what really matters is that when you have a relativistic quantum field theory with a particle of some given mass, it will have a preferred length scale and will thus not be invariant under scale transformations, hence not under conformal transformations... unless that mass is zero, in which case the Compton wavelength is undefined.
What is the meaning of the Compton wavelength?  Here's a rough explanation.   In quantum mechanics, to measure the position of a particle, you can shoot light (or some other kind of wave) at it.   To measure the position accurately, you need to use light with a short wavelength, and thus a lot of energy.  If you try to measure the position of a particle with mass $m$ more accurately than its Compton wavelength, you need to use an energy that exceeds $mc^2$.  This means that the collision is energetic enough to create another particle of the kind whose position you're trying to measure.  So, in relativistic quantum field theory, we should imagine any particle as part of a 'cloud' of 'virtual' particles which can become 'real' if you try to measure its position too accurately.  The size of this cloud is not sharply defined, but it's roughly the Compton wavelength.  So, the theory of this particle is not invariant under conformal transformations.
A: It's not clear from the post if you are talking about classical or quantum field theories. In QFT, conformal invariance implies scale invariance. If the theory has a mass $m$ then, as John explained, this defines a characteristic length, roughly $l=m^{-1}$ which has to be preserved by symmetries like scale transformations. So that means that the mass $m$ must satisfy
$$
\forall \lambda>0,\ \lambda m=m\ .
$$
The usual conclusion people infer from this requirement is that $m=0$, i.e., the mantra "only massless theories can be conformally invariant".
However, a bit in the same spirit as Carlo's answer, let me mention that this is not the only solution of the equation, namely, there is also $m=\infty$.
This corresponds to white noise, or the high temperature renormalization group fixed point. The probability school around Takeyuki Hida has shown that this is also conformally invariant (see references 107 and 108 mentioned on page 11 of my article "Towards three-dimensional conformal probability").
BTW, I am still waiting for an answer to this related MO question.

Edit: My understanding above of the OP's question was that only one mass $m$ is present, in which case the only possibilities are $m=0$ or $m=\infty$. I should add however, given the comments by Marcel and Logan, that another possibility is to have a continuous spectrum of masses present.
In fact this is the most common situation in CFT. For a field $\phi$ with scaling dimension $\Delta\in\left(\frac{d-2}{2},\frac{d}{2}\right)$ the Euclidean two-point function is up to constant factors
$$
\langle\phi(x)\phi(y)\rangle\sim\frac{1}{|x-y|^{2\Delta}}\sim
\int \frac{d\xi}{(2\pi)^d}e^{i\xi(x-y)}\frac{1}{|\xi|^{d-2\Delta}}
$$
$$
\sim\int \frac{d\xi}{(2\pi)^d}e^{i\xi(x-y)}\left(\int_0^{\infty}
m^{2\Delta-d+1}\frac{dm}{\xi^2+m^2}\right)
$$
$$
=
\int_0^{\infty}dm\ 
m^{2\Delta-d+1}\ \left(-\Delta+m^2\right)^{-1}(x,y)\ .
$$
The last formula is the explicit form of the Källén-Lehmann representation mentioned in Marcel's comment.
By a theorem of Pohlmeyer, CFTs which do not have this kind of continuous mass spectrum must be free. So in that sense, this third option is the most common one in the physics literature.
A: This may be a less "clear-cut-no" than suggested by the mantra "there can be no massive particles in a CFT because that would introduce a scale": Anatol Odzijewicz has constructed a CFT for massive particles, in which the mass is allowed to vary as a result of interactions:

The aim of this paper is the construction of a field theory for a
  massive conformal particle interacting with an external field. The
  classical conformal scalar massive particle is defined as a time
  spatially localized physical object with an energy and a momentum. Just
  as in relativistic mechanics the conformal particles will be divided
  here into particles, anti particles and tachyons. But contrary to the
  relativistic mechanics the mass of the particle may vary during its
  evolutions. Consequently the corresponding phase spaces will be the
  conformally homogeneous Hamiltonian symplectic manifolds of eight
  dimensions.

see A Conformal Holomorphic Field Theory
A: The usual mantra is usually formulated in the context of quantum field theory on Minkowski space. In a classical field theory, there is no particles and no mass for particles. In a quantum relativistic field theory, there are particles and it makes sense to talk about their mass (technically, a particle in quantum field theory is one of the irreducible components appearing discretely in the representation of the Poincaré group on the Hilbert space of states of the theory, and the mass of the particle is the corresponding value of the quadratic Casimir operator of the Poincaré Lie algebra).
In a relativistic theory, a length is the same thing as a time (length=c x times). In a quantum relativistic theory, it is also the same thing as the inverse of a mass or the inverse of an energy (energy = $\hbar$ /time). In particular, a non-zero mass scale implies a non-zero length scale and so a quantum relativistic theory with massive particles cannot be conformally invariant (technically, for an irreducible representation of the conformal group which  restricts to a discrete sum of irreducible representations of the Poincaré group, these irreducible representations of the Poincaré group are of zero mass).
A: So, you must be talking about quantum field theories, as mass is something we associated with particles, which are a quantum phenomenon in field theories.
In QFT, there are operators which create particles.  This means (neglecting various details, so please forgive the imprecision), that: 


*

*There is a distinguished element of the Hilbert space of states, $|vac\rangle$, the vacuum, which has no particles.

*There are operators $\phi_i(f)$ which, when applied to the vacuum, create a state $|(i,f)\rangle := \phi_i(f)|vac\rangle$ in which there is a particle of species $i$ with wavefunction $f$.


The mass of particles is measured by the asymptotic spatial decay rate of the inner products of these 1-particle states. 
$$\langle (i,f) | (i,g) \rangle = \int f(x) g(y) K_i(x,y)d^3x d^3 y$$
for some kernel $K_i$, which is sometimes called the 'correlation function'.  If we take $x$ and $y$ to be separated by some large distance $r$, we find that $K_i(x,y) \sim e^{-m_ir}$ in massive theories.  This can be taken as a definition of the mass $m_i$ of particles of type $i$.  This works on curved spacetimes, too.
In conformal quantum field theories, on the other hand, conformal invariance forces $K(x,y)$ to have power law decay $K(x,y) \sim r^{-\alpha}$. This is the basic incompatibility.  (In fact, conformal invariance completely determines the form of the two-point correlation functions, not just their asymptotics.  You can find a proof of this in almost any text on conformal QFT.)
