Anti-particle problem for Dirac sea According to the Dirac hole theory we know that Dirac sea is completely filled with negative energy, called vacuum. We will need $2mc^2$ or greater to get electron and a positron by incident photon.
Now my question is, why can't an electron  fall into the sea?
Is the reason angular momentum symmetry, and if a electron falls on the Dirac sea then conservation will be violated?
 A: No.
First of all, you should be careful with the Dirac sea, the way you describe it is not good.
When Dirac was thinking about his equation, I can imagine that the ideal picture for him would be: there is an equation that is Lorentz-invariant, admits positive-definite conserved density, etc, and gives only positive-energy states (=solutions).
However, it turns out that Dirac equation for each positive-energy state has a corresponding negative-energy state. One could therefore imagine an electron to 'fall' down and occupy a negative-energy state. This is, however, physically meaningless, so there is this idea of Dirac sea, which states:
In a vacuum state all negative-energy states are occupied by unseen electrons.
By the Pauli exclusion principle, this means that if you now have a real positive-energy electron added to this vacuum, it cannot fall down to negative states, because there is no space left. These unseen negative-energy electrons are called the Dirac sea.
Now, an electron from the Dirac sea can pop up to some positive energy state, if we provide it with enough energy (and here you are right, the gap is $2mc^2$). Then there are now a positive-energy electron, and a hole in the Dirac sea. This hole behaves like a real particle, and is identified with a positron. This is the process of $e^+e^-$ production.
So, when there is now a vacant state in the Dirac sea, an electron can fall down to this state. This looks like a simultaneous vanishing of an electron and a positron (the vacant place) and is called $e^+e^-$ annihilation. The energy the electron loses is released in form of two photons.
The property of these Dirac-sea electrons being unseen is also natural, since if an interaction of something with a Dirac-sea electron is non-trivial (leads to observable effects), it should change the state of the electron. However, all but a few negative energy states are occupied. So this electron has no place to go than either to positive energy state ($e^+e^-$ production) or a vacant state (and this is observed as an interaction of the positron). Nevertheless, this picture is rather hand-waving, and in QFT the problem is resolved by a mere renaming of some operators, which is much more formal and convincing.
