When the electromagnetic waves propagate without energy losses, e.g. in the vacuum, it is easy to prove that the total energy is conserved. See e.g. Section 1.8 here.
In fact, not only the total energy is conserved. The energy is conserved locally, via the continuity equation
$$\frac{\partial \rho_{\rm energy}}{\partial t}+\nabla\cdot \vec J = 0$$
This says that whenever the energy decreases from a small volume $dV$, it is accompanied by the flow of the same energy through the boundary of the small volume $dV$ and the current $\vec J$ ensures that the energy will increase elsewhere. The continuity equation above is easily proven if one substitutes the right expressions for the energy density and the Poynting vector:
$$ \rho_{\rm energy} = \frac{1}{2}\left(\epsilon_0 E^2+ \frac{B^2}{\mu_0} \right), \quad \vec J = \vec E\times \vec H $$
After the substitution, the left hand side of the continuity equation becomes a combination of multiples of Maxwell's equations and their derivatives: it is zero.
These considerations work even in the presence of reflective surfaces, e.g. metals one uses to build a double slit experiment. It follows that if an electromagnetic pulse has some energy at the beginning, the total energy obtained as the integral $\int d^3x \rho_{\rm energy}$ will be the same at the end of the experiment regardless of the detailed arrangement of the interference experiment.
If there are interference minima, they are always accompanied by interference maxima, too. The conservation law we have proved above guarantees that. In fact, one may trace via the energy density and the current, the Poynting vector, how the energy gets transferred from the minima towards the maxima.
Imagine that at the beginning, we have two packets of a certain cross section area which will be kept fixed and the only nonzero component of $\vec E$ goes like $\exp(ik_1 x)$ (and is localized within a rectangle in the $yz$ plane). It interferes with another packet that goes like $\exp(ik_2 x)$. Because the absolute value is the same, the energy density proportional $|E|^2$ is $x$-independent in both initial waves.
When they interfere, we get
$$\exp(ik_1 x)+\exp(ik_2 x) = \exp(ik_1 x) (1 + \exp(i(k_2-k_1)x) $$
The overall phase is irrelevant. The second term may be written as
$$ 1 + \exp(i(k_2-k_1)x = 2\cos ((k_2-k_1)x/2) \exp(i(k_2-k_1)x/2)$$
The final phase (exponential) may be ignored again as it doesn't affect the absolute value. You see that the interfered wave composed of the two ordinary waves goes like
$$ 2\cos ((k_2-k_1)x/2) $$
and its square goes like $4\cos^2(\phi)$ with the same argument. Now, the funny thing about the squared cosine is that the average value over the space is $1/2$ because $\cos^2\phi$ harmonically oscillates between $0$ and $1$. So the average value of $4\cos^2\phi$ is $2$, exactly what is expected from adding the energy of two initial beams each of which has the unit energy density in the same normalization. (The total energy should be multiplied by $A_{yz} L_x\epsilon_0/2$: the usual factor of $1/2$, permittivity, the area in the $yz$-plane, and the length of the packet in the $x$-direction, but these factors are the same for initial and final states.)
Finally, let me add a few words intuitively explaining why you can't arrange an experiment that would only have interference minima (or only interference maxima, if you wanted to double the energy instead of destroying it – which could be more useful). To make the interference purely destructive everywhere, the initial interfering beams would have to have highly synchronized phases pretty much at every place of the photographic plate (or strictly). But that's only possible if the beams are coming from nearly the same direction. But if they're coming from (nearly) the same direction, they couldn't have been split just a short moment earlier, so it couldn't have been an experiment with the interference of two independent beams. The beams could have been independent and separated a longer time before that. But if the beams started a longer time before that, they would still spread to a larger area on the photographic plate and in this larger area, the phases from the two beams would again refuse to be synchronized and somewhere on the plate, you would find both minima and maxima, anyway.
The argument from the previous paragraph has simple interpretation in the analogous problem of quantum mechanics. If there are two wave packets of the wave function for the same particle that are spatially isolated and ready to interfere, these two terms $\psi_1,\psi_2$ in the wave function are orthogonal to each other because their supports are non-overlapping. The evolution of the wave functions in quantum mechanics is "unitary" so it preserves the inner products. So whatever evolves out of $\psi_1,\psi_2$ will be orthogonal to each other, too, even if the evolved wave packets are no longer spatially non-overlapping. But this orthogonality is exactly the condition for $\int|\psi_1+\psi_2|^2$ to have no mixed terms and be simply equal to $\int|\psi_1|^2+|\psi_2|^2$. The case of classical Maxwell's equations has a different interpretation – it's the energy density and not the probability density – but it is mathematically analogous. The properly defined "orthogonality" between the two packets is guaranteed by the evolution and it is equivalent to the condition that the total strength of the destructive interference is the same as the total strength of the constructive interference.
