As mentioned in the other answers the etendue theorem rules this out for a system of mirrors and lenses. However, I think it's important to note that simple thermodynamic arguments are insufficient for the reasons given below.
I will answer the question using mirrors rather than lenses as it makes the physics clearer. Suppose we have a massive mirrored cavity at 0K:
At some time we switch on the sun:
Now the cavity will slowly start to fill with radiation. Once the radiation density in the cavity has reached a certain level (which depends only on the sun's temperature) the sun will absorb just as much radiation as it emits and thermal equilibrium will be reached. Note that before equilibrium is reached heat is transfered from the sun into the cavity.
Now suppose we add another black body at the other end of the cavity, initially at lower temperature than the sun:
Is it possible for the second black body to reach a temperature higher than the sun? Obviously not, for once it's temperature reaches that of the sun it will also absorb and emit radiation to/from the cavity at the same rate, and no more heat will be transfered (for simplicity, we are assuming that the sun's temperature is fixed).
However, this alone does not rule out the possibility to heat a second black body to a temperature higher than the sun's surface only by focussing the solar radiation. To understand why, it is useful to consider Anna's example of the LHC in more detail.
Assuming the LHC was powered only by solar energy (e.g. converted into electricity by photovoltic cells), this is what is happening. Heat $Q_{12}$ is transfered from the sun (temperature $T_1$) to the solar cells (temperature $T_2<T_1$) which act as heat engines converting part of the transfered heat into work $W<Q_{12}$. This work then powers a heat pump to transfer heat $Q_{23}$ to the quark-gluon plasma in the LHC (temperature $T_3>T_2$). The second law of thermodynamics puts some limits on the ratios of $Q_{12}:W$ and $W:Q_{23}$, but there is no problem with heat from the sun being transfered to the hotter plasma, due to the lower temperature of the intermediate system (the photovoltic cells).
So, we can use the sun's radiation to increase the temperature of a black body above that of the sun provided there is some intermediate system at lower temperature. In the OP's original question the space itself can in principle play the role of the intermediate system.
Consider the mirror cavity example again, but this time instead of allowing the sun to equilibriate with the cavity before adding the second body we add both to the 0K cavity at the same time, and assume that the temperature of the second body is equal to the temperature of the sun. Initially, the space inbetween the two black bodies is at a lower temperature to either and (as far as the second law of thermodynamics is concerned) there can be a net heat transfer from the sun to the second body provided there is a corresponding heat transfer from the sun into the cavity.
As mentioned above, the optics doesn't allow this, but I can't see any reason to rule it out on purely thermodynamic grounds.
Response to comments by Hypnosifl
Hypnosifl suggests that the non-equilibrium result can be derived from the equilibrium one with 'minimal assumptions' about the nature of the radiation. The gist of the argument (if I understand it correctly) is that since the radiation incident on the second black-body surface depends only on the source (the sun), this should be the same whether at equilibrium or not. I do not think this argument is sufficient for the following reason. In order for the the radiation field to ever reach equilibrium with the sun it is necessary to enclose it in a cavity of some kind (say a perfectly mirrored cavity for simplicity), in which case the radiation incident on the second body at equilibrium will include all sorts of complicated reflections of the cavity walls. Thus, until equilibrium is reached the incident radiation will be time dependent. (For example, suppose the second body is the Earth. Then the first radiation from the sun will reach the Earth after about eight minutes, and this will be radiation only directly from the sun. Some time later rays reflected once of the cavity will start to arrive, then double reflections, and so on.) Thus we cannot conclude that the luminance is the same in the non-equilibrium case as in the equilibrium case. You could maybe argue that it is impossible for the incident flux to decrease as a function of time (which is sufficient), but this is a more complicated argument.
[Edit:] It turns out that it is possible for the incident flux to decrease as a function of time after all, as the following example shows. Pass the radiation from the sun through a collimator into a large box. Then pass the collimated beam through a prism to separate different frequencies of light. Choose a particular frequency and pass that part of the separated light beam through another collimator. You now have a (reasonably) coherent beam of light. Now send this beam of light into a Michelson-Morley style interferometer with one arm significantly shorter than the other and place the test surface where the 'scope of the interferometer is. Light travelling the shorter path will reach the test surface first, giving a uniform intensity across the test surface. When the second beam arrives inteference fringes will be created, meaning that certain parts of the surface will see a decrease in intensity (while others will see an increase). Thus the argument at the end of the previous paragraph will not work. In any case, it is becoming clear that the thermodynamic argument requires a significant amount of extra work to complete it. [End edit]
Last night I thought of a similar argument from the equilibrium case to the non-equilibrium case, based upon the fact that at equilibrium thermodynamics tells us that the energy density of radiation is the same everywhere in the cavity (as long as you are in vacuum) and independent of the shape of the cavity and the direction of the radiation. By imagining how the radiation flows from one part of the cavity to another, this result seems to rule out the possibility of concentrating radiation from a uniform source (i.e. a source which emits equally in all directions) to an intensity higher than at the surface of the source in another part of the cavity, from which the non-equilibrium result follows. However, (i) I haven't made this argument rigorous and (ii) I suspect that the content of any rigorous argument along these lines would be similar to the conservation of etendue (i.e. it looks like some kind of conservation of volume in phase space). Thus this could hardly be called a 'minimal assumption' either.
