From the principles of quantum field theory, there's an argument you can make for antiparticles as follows (this is following the presentation in Weinberg's textbook, but I omit some details and might get some factors of 2 or -1 wrong).
First of all, you ask how you can make a unitary quantum mechanical theory which is relativistic. It turns out that this is possible if every observable of the theory is written in terms of causal quantum fields, which are operators satisfying
<br />
[\phi_1(x,t),\phi_2(x',t')] = 0 \qquad (x - x')^2 > (t - t')^2<br />
for all pairs of fields and any composite you can make out of them. This ensures that the time-ordering of measurements outside of each other's light cones doesn't matter (which is already a nice property to have, considering the no-communication theorem).
Now you expand your fields in terms of creation and annihilation operators, but in order to satisfy the above relation you need to do so in a particular way. You find that the fields need to have the form (schematically)
<br />
\phi(x,t) \sim \int d\omega \, d^d k \left[ e^{- i \omega t + i p x}c(p,\omega) + e^{i \omega t + i px} c^{\dagger}(p,\omega) \right]<br />
where c and c^{\dagger} destroy and create operators respectively. Note the extra sign in front of the time-dependence of the second term: this will eventually convince Feynman that antiparticles are particles moving backwards in time, but it exists even without antiparticles, and is just required by (1) expanding fields linearly in creation/destruction operators and (2) enforcing the commutation relation above.
Ok, now what about if you have a conserved charge, as we do in our universe? In this case we not only want our Hamiltonian to be a Lorentz scalar, we also want it to commute with some operator Q. If the particle created by c^{\dagger} has charge q, then
<br />
[Q,c(p,\omega)] = - q \, c(p,\omega)<br />
<br />
[Q,c^{\dagger}(p,\omega)] = q \, c^{\dagger}(p,\omega)<br />
But now the commutation relation of Q with \phi is all messed up! This is a nightmare, since we spent so much time constructing the \phi fields so that we can write our Hamiltonian as a function of them, but now we want [Q,H] = 0 which can't be done with the fields as-is.
Here, Weinberg argues that you need to fix this by introducing a new species of particle with charge -q (an "antiparticle"), and writing
<br />
\phi(x,t) \sim \int d\omega \, d^d k \left[ e^{- i \omega t + i p x}c(p,\omega) + e^{i \omega t + i px} a^{\dagger}(p,\omega) \right]<br />
where now a^{\dagger}(p,\omega) creates an antiparticle. Now you have [Q,\phi] = q\, \phi and [Q,\phi^{\dagger}] = -q\, \phi^{\dagger}, and enforcing [Q,H] = 0 simply boils down to each term having an equal number of \phi's and \phi^{\dagger}'s.
(There are a few assumptions here, like that we need (and have) an expansion in terms of creation/annihilation operators.)