The correct normalization factor is $$ N = \frac{1}{\sqrt{2}}.$$
To see this, note that you can write your wave-function in ket notation as
$$\psi(x) = \langle x | a \rangle + \langle x | -a \rangle \equiv \langle x | \psi \rangle, $$
where we have used the usual basis for the (one dimensional) position representation, with normalization
$$ \langle x | y \rangle = \delta(x-y),$$
and we have defined the state $|\psi\rangle$ as the state corresponding to your wave function:
$$| \psi \rangle = | a \rangle + |-a \rangle.$$
Your question now translates into: what is $\langle \psi | \psi \rangle$?
The answer is readily obtained: this is a sum of two orthogonal states (assuming $a\neq0$), hence
$$ \langle \psi | \psi \rangle = \langle a | a \rangle + \langle -a | -a \rangle = 2$$
and
$$ | \psi \rangle = \frac{1}{\sqrt{2}} ( |a\rangle + |-a\rangle),$$
$$ \psi(x) = \frac{1}{\sqrt{2}} ( \delta(x-a)+ \delta(x+a)).$$
Note that if you try to compute $\langle a | a \rangle$ as an integral you get $\infty$:
$$ \langle a | a \rangle = \int dx \delta(x-a)^2 \approx \delta(0) \approx \infty $$
where the $\approx$ is because we are being mathematically sloppy here, see this discussion of Qmechanic about it.
This is due to the use of a continuous base. You can always think of it as the limiting case of a discrete base, taking case this way of the mathematical difficulties.
