Can the magnitude of a vector sum ever equal the sum of the magnitudes?

Yes. This occurs exactly when one vector is a nonnegative scalar multiple of the other.

Working from your point of view you want, for nonzero $\vec\alpha$ and $\vec\beta$, $$ (\|\vec\alpha\|+\|\vec\beta\|)^2=\|\vec\alpha+\vec\beta\|^2=\|\vec\alpha\|^2+\|\vec\beta\|^2-2\|\vec\alpha\|\, \|\vec\beta\|\,\cos(\pi-\theta). $$ So you need $\cos(\pi-\theta)=1$, which is exactly $\pi=\theta$. So the two vectors are colinear.

In $\mathbb{R}^n$, we have that your equality implies $$\langle a+b,a+b \rangle=(\Vert a\Vert+\Vert b \Vert)^2 ,$$ which, using the bilinearity of the inner product, yields $$\Vert a \Vert^2+\Vert b \Vert^2 +2\langle a,b \rangle=\Vert a\Vert^2+\Vert b \Vert ^2 +2\Vert a \Vert \Vert b \Vert$$ $$\implies \langle a,b \rangle= \Vert a \Vert \Vert b \Vert.$$ And equality on Cauchy-Schwarz only holds if both vectors are linearly dependent. (Note that since there is no modulus on the inner product on the left side, not only they must be linearly dependent, but also differ by a positive scaling).

As a sidenote, the technique above holds for any inner product space. One might be tempted to extend it to any normed space, but the result isn't true. As an example, one can take $L^1([0,1])$ and $a=I_{[0,1/2]}$, $b=I_{[1/2,1]}$, where $I_A$ is the indicator function on $A$.

Recall that $\|\vec x\|^2 = \vec x \cdot \vec x$.

If one vector is zero then the equation is trivially true, so suppose they are nonzero. If you square both sides you get $$ \| \vec \alpha \|^2 + 2(\vec \alpha \cdot \vec \beta) + \| \vec \beta \|^2 = \| \vec \alpha \|^2 + 2\|\vec \alpha\| \|\vec \beta\| + \| \vec \beta \|^2$$ so that $\vec \alpha \cdot \vec \beta = \|\vec \alpha\| \|\vec \beta\|$. This can only happen if the two vectors point in the same direction. (To see this, look at the Cauchy-Schwarz inequality says, or compute the angle $$\theta = \arccos \frac{\vec \alpha \cdot \vec \beta}{\|\vec \alpha\| \|\vec \beta\|}$$ between the two vectors.)