Let $\mathcal{X} = \left\{ \mathbf{x}_{i} \right\}_{i=1}^N$ be a set of vectors where $\mathbf{x}_{i} \in \mathbb{R}^{d}$.
I know that the minimizer of $\arg \min_{\mathbb{y}} \sum_{i = 1}^{N} \left\| \mathbf{y} - \mathbf{x}_{i} \right\|_2^2$ is the mean of the set.
I also can calculate it per component, so I can calculate $d$ times the mean of $N$ scalars.
I know the minimizer of $\arg \min_{\mathbb{y}} \sum_{i = 1}^{N} \left\| \mathbf{y} - \mathbf{x}_{i} \right\|_1$ is the median.
Can I calculate it in the same way? By calculating $d$ medians of $N$ scalars (Components of each vector)?
You can actually decompose the problem by coordinate:
$$ \|y - x_i\|_1 = \sum_{j = 1}^d |y_j - (x_i)_j| \implies \sum_{i} \|y - x_i\|_1 = \sum_{i, j} |y_j - (x_i)_j| = \sum_{j = 1}^d \left( \sum_{i = 1}^n |y_j - (x_i)_j| \right) $$
Therefore, you have
$$ \mathrm{min}_{y} \sum_{i = 1}^n \|y - x_i\|_1 = \left( \mathrm{min}_{y_1} \sum_{i = 1}^n |y_1 - (x_i)_1| \right) + \dots + \left( \mathrm{min}_{y_d} \sum_{i = 1}^n |y_d - (x_i)_d| \right). $$
Each of the one-dimensional problems is solved by the median across the corresponding coordinate.