For the nearest neighbor hopping model on the Kagome lattice, there is a flat band among the three energy bands. 
Is there some reason, such as symmetry or the special structure of the model, to ensure the existance of a flat band?
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Yes, there is a structural reason for the existence of flat band on the Kagome lattice. This is related to the wave function localization due to the destructive interference on the lattice.
The flat band has many physical interpretations. In the momentum space, looking at the dispersion relation, a flat band means the effective mass of the particle is infinite, i.e. the particle is "super heavy" such that it can not move and therefore not dispersive. In the real space, it means the hopping of the particle between different regions is effectively turned off, such that the particle is localized and just sit there one-by-one, like the atomic limit in solid-state physics. The energy levels of all states are degenerated, and there is no hybridization, no energy band broadening, and hence the band has to be flat.
Given the above understanding of the flat band physics, the key point is to explain why the particle effectively can not move/hop on the Kagome lattice. This can be explained by the destructive interference of the following wave function, which is depicted in the following figure with red site = positive weight, blue site = negative weight (of the same amplitude as red), and white site = no weight.
This is a wave function localized in the hexagon, and the particle is actually circulating around the hexagon with the momentum pi. Due to the special structure of the Kagome lattice, even if the nearest neighbor hopping is there, this wave function still can not propagate out of the hexagon. Because the only way to escape the hexagon is to first hop to the white sites on the corner, but the weight can be transferred from both the red and the blue sites (that share the same triangle with the white) with equal probability, and their weights are opposite in sign and same in amplitude, so they will be canceled out exactly on the white site. The result is that this hexagon-ring wave function can never enter the white site, and hence no hope to travel to the whole lattice outside.
Such a destructive interference induced self-localization phenomenon happens on each hexagon. If any particle is trapped in such a hexagon-ring state, it will never be able to get out of the hexagon. Therefore all these particles are localized, with no kinetic energy, so they constitute the states in the flat band.
However, strictly speaking, different hexagon-ring states are not orthogonal to each other, which can be seen from the fact that the nearest hexagons still overlap with each other on one corner-sharing site. But this is not a problem. Although not orthogonal, these states are still linearly independent, and there are (roughly) as many hexagons as the number of states in the Brillouin zone. As these hexagon ring states are all degenerated in energy, we are free to orthogonalize them. After the orthogonalization, we will obtain the basis states in the flat band.
In conclusion, there is a structural reason for the flat band on the Kagome lattice. Self-localization under destructive interference is the key to the explanation. In fact, all the flat bands in condensed matter systems are explained in the same manner, including the famous example of the Landau level for charged particles in the magnetic field, where the destructive interference is due to the Berry phase, which, however, is topological other than structural.
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$\begingroup$ Thanks a lot! So the flat bands of this kind never have non-trivial Chern number, since there are Wannier functions which are well localized. Is that right? Therefore another question arises: how to explain the flat band with non-trivial Chern number following the same strategy? $\endgroup$– TengenCommented Dec 18, 2013 at 12:05
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$\begingroup$ @Tengen In general, the flat band is not associated with non-trivial Chern number. For example, the Kagome flat band has no Chern number, and in fact the Chern number is not even well-defined due to the gapless character at the band touching point. For the gapped flat bands, many of them do have non-trivial Chern numbers, but I do not see any general prove that there should be non-trivial Chern number. $\endgroup$ Commented Dec 18, 2013 at 18:58
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1$\begingroup$ Thouless has a theorem stating that the Wannier functions are exponentially localized if and only if the band has zero Chern number. The wave functions constructed above are of course localized, even though they are not exactly Wannier functions. Therefore I guess all the flat bands of this kind have zero Chern number. Does the destructive interference picture apply to bands with nonzero Chern number? $\endgroup$– TengenCommented Dec 19, 2013 at 2:55
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2$\begingroup$ No, your example is wrong. Because these states are neither complete nor linearly independent. Please check Thouless' paper $\endgroup$– TengenCommented Dec 19, 2013 at 10:56
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2$\begingroup$ It is true that a nonzero Chern number implies that there is not a complete set of localized Wannier functions. The reason is simple: Wannier functions are related to $\psi_\mathbf{k}$ by a Fourier transform, so if the former is integrable (which localized functions are), then the latter is smooth in $\mathbf{k}$, but a non-zero value of the Chern number is only possible if $\psi_\mathbf{k}$ is not differentiable for some $\mathbf{k}$. $\endgroup$ Commented Apr 19, 2016 at 14:48