Anticommutation relations and bispinor field In a case of free Dirac field we have
$$
\hat {H} = \int \epsilon_{\mathbf p}\left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) - \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p ) \right)d^{3}\mathbf p,
$$
$$
\hat {\mathbf P} = \int \mathbf p \left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) - \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p )  \right)d^{3}\mathbf p,
$$
$$
\hat {Q} = \int \left( \hat {a}^{+}_{s}(\mathbf p )\hat{a}_{s}(\mathbf p ) + \hat{b}_{s}(\mathbf p )\hat{b}^{+}_{s}( \mathbf p )\right)d^{3}\mathbf p.
$$
So, if an operator $\hat {b}^{+}$ act on energy vector, it will decrease the value of an energy. This is commonly referred to as the fact that it creates an antiparticle with negative energy, because $\hat {b}^{+}$ is interpreted as creation operator (only by analogy with the scalar field, if I understand correctly). The solution to this problem is the postulation of anti-commutation relations between operators (in addition, the summary energy of the field begin to be positive definite quantity). 
But why don't we call $\hat {b}$ the creation operator, and $\hat {b}^{+}$ the destruction operator? Then, requiring the positivity condition for the integrand (if it is possible), we'll get physically correctly result without postulation of anticommutation relations.
Where did I make the mistake?
 A: We call $b^\dagger$ the creation operator exactly because it increases the energy of the state (by adding a quantum), in contradiction with your incorrect first sentence. On the contrary, $b$ is called the annihilation operator because it decreases the energy and because, when it acts on the vacuum $|0\rangle$, it actually produces zero.
You made a sign mistake while writing the first sentence. The first equation, which you wrote correctly, may be rewritten as
$$\hat {H} = E_0+ \int \epsilon_{\mathbf p}\left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) + \hat {b}_{s}^+(\mathbf p )\hat {b}_{s}(\mathbf p ) \right)d^{3}\mathbf p$$
Note that the sign in front of $b^\dagger b$ was switched to minus exactly because they (almost) anticommute with each other. The anticommutator was added to the irrelevant $c$-number term $E_0$ that I wrote in front of the integral and that we ultimately set to zero.
Note that in my form, both $a^\dagger$ and $b^\dagger$ create positive-energy quanta (with opposite charge, positrons and electrons), and they generally enter symmetrically. This treatment that sees fermions and antifermions symmetrically may be obtained by renaming the negative-energy $a^\dagger,a$ as $b,b^\dagger$ – by using the word "positron" for a hole in the mostly filled Dirac sea of negative-energy electron states. This is possible because there's a symmetry between $a,a^\dagger$ – their anticommutator is symmetric in them, much like the occupation numbers $0,1$ for a fermionic state are symmetric with respect to the value $1/2$. So it's possible to rename empty vs occupied, exchange the notation for creation vs annihilation operators, and end in a picture in which the creation operators always increase the energy.
