= Solution
With a vacuum annihilated by all $a_{\mathbf p}^s,b_{\mathbf p}^s$, define the <particle and antiparticle occupation operators of a Dirac field>
$$
N_e=\sum_s\int\frac{d^3p}{(2\pi)^3}a_{\mathbf p}^{s\dagger}a_{\mathbf p}^s,\qquad
N_{\bar e}=\sum_s\int\frac{d^3p}{(2\pi)^3}b_{\mathbf p}^{s\dagger}b_{\mathbf p}^s.
$$
Each counts its corresponding occupied modes. The total positive particle count is
$$
\boxed{N_{\rm total}=N_e+N_{\bar e}.}
$$
Commuting it through either creation operator gives $[N_{\rm total},a^\dagger]=a^\dagger$ and $[N_{\rm total},b^\dagger]=b^\dagger$, so both kinds of one-particle states have eigenvalue one. In a normalized discrete mode, $(a^\dagger a)^2=a^\dagger a$ by the anticommutation relations, so its occupation eigenvalues are $0,1$.
For a <Dirac field> there is also a distinct conserved signed fermion number. Inserting the mode expansion into the <normal-ordered> local density gives
$$
\boxed{Q_D=\int d^3x:\psi^\dagger\psi:
=N_e-N_{\bar e}.}
$$
Spatial integration makes the same-sector momenta equal. The cross terms vanish by $u^\dagger(\mathbf p)v(-\mathbf p)=0$; <normal ordering> of $bb^\dagger$ supplies the minus sign for the <antiparticles>. Thus the density integral is not the total positive particle count. It generates the global phase symmetry, and for <electron> charge $-e$ the electric charge is $Q_{\rm em}=-eQ_D$.
The free <normal-ordered> <Hamiltonian> is
$$
H=\sum_s\int\frac{d^3p}{(2\pi)^3}E_{\mathbf p}
\left(a_{\mathbf p}^{s\dagger}a_{\mathbf p}^s+
b_{\mathbf p}^{s\dagger}b_{\mathbf p}^s\right),
$$
so $N_e,N_{\bar e},N_{\rm total}$ and $Q_D$ are conserved in the free theory. In an interacting theory, pair creation can change the total count while preserving the signed charge. This distinguishes a positive occupation <number operator> from the charge behind <Dirac fermion number conservation>.
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