Let the two non-zero vectors be $\vec{A}$ and $\vec{B}$, and let $\theta$ be the angle between them.

The magnitude of the sum of the two vectors is given by: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos\theta}$

The magnitude of the difference of the two vectors is given by: $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos\theta}$

According to the problem statement, the magnitude of their sum is equal to the magnitude of their difference: $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$

Squaring both sides to remove the square roots: $(A^2 + B^2 + 2AB \cos\theta) = (A^2 + B^2 - 2AB \cos\theta)$

Now, simplify the equation: $A^2 + B^2 + 2AB \cos\theta = A^2 + B^2 - 2AB \cos\theta$

Subtract $A^2 + B^2$ from both sides: $2AB \cos\theta = -2AB \cos\theta$

Add $2AB \cos\theta$ to both sides: $4AB \cos\theta = 0$

Since $\vec{A}$ and $\vec{B}$ are non-zero vectors, their magnitudes $A$ and $B$ are not zero. Therefore, for the product $4AB \cos\theta$ to be zero, $\cos\theta$ must be zero. $\cos\theta = 0$

The angle $\theta$ for which $\cos\theta = 0$ is $90^\circ$ (or $\pi/2$ radians).

Let's analyze why other options are incorrect:

• If $\theta = 0^\circ$, then $\cos\theta = 1$. The equation becomes $4AB(1) = 0$, which implies $A=0$ or $B=0$, contradicting the condition that they are non-zero vectors.
• If $\theta = 45^\circ$, then $\cos\theta = 1/\sqrt{2}$. The equation becomes $4AB(1/\sqrt{2}) = 0$, which again implies $A=0$ or $B=0$.
• If $\theta = 180^\circ$, then $\cos\theta = -1$. The equation becomes $4AB(-1) = 0$, which also implies $A=0$ or $B=0$.

Thus, the only valid angle for non-zero vectors is $90^\circ$.

The final answer is $\boxed{90^\circ}$.