Let $z \in \mathbb{C}^m.$ Then $\| \sum_{j} z_j \|_2 = \sum_j \| z_j \|_2 \Longleftrightarrow z_j = \alpha_j z_1$ for some $ \alpha_j>0.$ Let $ z = ( z_1, z_2, \dots, z_m) \in \mathbb{C}^m$ and $z_j>0$ for all $ 1 \leq j \leq m.$ Then
$\| \sum_{j} z_j \|_2 = \sum_j \| z_j \|_2$ if and only if $ z_j = \alpha_j z_1$ for some $ \alpha_j>0.$
This side is clear:  If $ z_j = \alpha_j z_1$ for $ \alpha_j>0,$ then $ \| \sum_{j} z_j \|_2 = \| z_1 + \alpha_2 z_1 + \dots + \alpha_m z_1 \|_2 = \| ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m )z_1\|_2 = ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m ) \|z_1\|_2 = \|z_1\|_2 + \|z_2\|_2 + \dots + \| z_m\|_2 = \sum_{j} \|z_j\|_2.$
For the other side of the proof, I know that I need to use the case of equality in the Cauchy-Bunyakovskii-Schwarz(CBS) Inequality, which says that:
For all $x,y \in \mathbb{C}^{m \times 1 }$ we have  \begin{align} |x^ * y| = \|x\| \|y\| \text{, where $x^*$ is the conjugate transpose of x,} \end{align} if and only if $ y = \alpha x$ for $ \alpha = \dfrac{x^ * y}{x^ * x}.$
Any directions or help with the proof?
 A: Here is an answer for the newly formulated question. I will actually assume that $z_j\in\mathbb{C}^k$, as this does not change the proof.
Let us start with $m=2$. Then we have $\|z_1+z_2\|_2=\|z_1\|_2+\|z_2\|_2$; using that $\|x\|_2=x^*x$ and squaring we get
$$
\|z_1\|_2+\|z_2\|_2+2\text{Re }z_1^*z_2=\|z_1\|_2+\|z_2\|_2+2\|z_1\|_2\|z_2\|_2.
$$
So
$$
\text{Re }z_1^*z_2=\|z_1\|_2\|z_2\|_2.
$$
But then
$$
\|z_1\|_2\|z_2\|_2=\text{Re }z_1^*z_2\leq|z_1^*z_2|\leq\|z_1\|_2\|z_2\|_2,
$$
which implies that 
$$
z_1^*z_2=\|z_1\|_2\|z_2\|_2.
$$
This is equality in the CBS inequality, so $z_2=\alpha_2\,z_1$, with $\alpha_2=z_1^*z_2/\|z_1\|_2^2\geq0$. Note that if $z_1,z_2$ are complex numbers, then $\alpha_2>0$.
Now we proceed by induction. If we have the equality for $h+1$ vectors, then
$$
\sum_1^{h+1}\|z_j\|_2=\|\sum_1^{h+1}z_j\|_2\leq\|\sum_1^{h}z_j\|_2+\|z_{h+1}\|_2\leq\sum_1^{h+1}\|z_j\|_2.
$$
Then we get equality for the first $h$ vectors and we can apply the inductive hypothesis to get $z_j=\alpha_jz_1$ for $j=1,\cdots,h$. Now $\sum_1^hz_j=(\sum_1^h\alpha_j)z_1$ and we can apply the $m=2$ case to get the final result.
A: We assume  $$\left|\sum_{k=1}^mz_k\right|=\sum_{k=1}^m|z_k|.
$$
 We consider two cases. The first one is when $\sum_{k=1}^mz_k\geq0$. Then we have
  $$\sum_{k=1}^mz_k=\sum_{k=1}^m|z_k|.
$$ Then
  $$0=\sum_{k=1}^m|z_k|-z_k.
$$
As every term in the sum is nonnegative, we get that $|z_k|=z_k$ for all $k$, so $z_k>0$ for all $k$. Now take $\alpha_k=z_k/z_1$.
The second case, when $\sum_{k=1}^mz_k<0$, is treated similarly and we get $|z_k|+z_k=0$ for all $k$. Now $z_k=-|z_k|<0$ for all $k$. So we can define $\alpha_k=z_k/z_1>0$ (quotient of two negative numbers).
This problem is definitely more interesting when $z_1,\ldots,z_m$ are vectors. Then the $\|\cdot\|_2$ notation is then justified, and the proof indeed requires the use of the case of equality in the CBS inequality.
