In quantum theory of elementary particles (in a sense of irreducible representation of Poincare group with mass $m$ and spin (helicity) $s$) if some operator $\hat{Q}$ of internal symmetry commutes (like electric charge charge) with Hamiltonian $\hat{H}$ of given field theory, there must be
$$
[\hat{Q},\hat{\varphi}^{\dagger}_{A}(\mathbf p)] | \rangle = q_{A}\hat{\varphi}^{\dagger}_{A}(\mathbf p), \quad [\hat{Q},\hat{\varphi}_{A}(\mathbf p)] | \rangle = -q_{A}\hat{\varphi}_{A}(\mathbf p).
$$
Here $\hat{\varphi}^{\dagger}(x)$ is a field which creates particles while $\hat{\varphi}(x)$ is a field which creates antiparticles.
So you can see two facts: the particle has charge which is equal to the charge of corresponding antiparticle with minus sign; the setting of sign of particle's charge is formal. We can set the positron charge as negative, the physics doesn't change.
Charge is only the measure of interaction. Let's see this on the simple example.
We know that there is interaction between two electrons which can't be described as gravitational (they are repelled). Also we know that proton and electron are attracted (this interaction also can't be described as gravitational because it's stronger than it). Then we know that this interaction force at the simplest case coinsides with $\frac{1}{r^{2}}$ law, and the measure of interaction is constant in time. We need to build the theory with these properties. The charge of given particle, as (for simplicity) the mass in case of gravitational interaction, says us about its taking part in interaction.
"positive" and "negative" are (historical) conventions, just like "rose" and "tulip". If we would swap these terms, we could still tell the charges (and the flowers) apart. In that sense there is no absolute meaning to either. The much more important fact is that there are two different polarities of charges which exist in equal numbers.
– CuriousOne Oct 19 '14 at 19:27