In a metric space $X$, let $I(x,y)$ denote the metric interval between two points, i.e. $I(x,y):=\{z:d(z,x)+d(z,y)=d(x,y)\}$. Given a triplet $(x,y,z)$, we say that $w$ is a median of the triplet if $w\in I(x,y)\cap I(x,z)\cap I(y,z)$. By some metric geometry, it is known that every 1-injective Banach space admits medians (see below). Is this a complete characterization? If not, what's an example of such a Banach space?
Addendum: Proof of "every 1-injective space admits medians".
It is well known that a Banach space is 1-injective iff it is an injective metric space (also known as tight or hyperconvex metric space). Thus the result follows if we prove that every injective metric space admits medians. To do that, let $x,y,z\in X$ ($X$ injective metric space). If one of $(x|y)_z, (x|z)_y, (y|z)_x$ is zero (here $(x|y)_z$ denotes the Gromov product), then the base point is already the median. If not, we construct a metric space $\{x,y,z,a\}$ with $d(a,x)=(y|z)_x$ and similarly for the other points and then use injectivity to get a Lipschitz embedding of $\{x,y,z,a\}$ in $X$. The image of $a$ in $X$ is the desidered median.
2° Addendum: to give some idea of the problem, here's an example of a Banach space with medians and a sketch of how to find them: $\ell^\infty_2$. In it, the metric segment between two points $x,y$ consists of the (possibly degenerate) rectangle with two sides parallel to the diagonal having $x$ and $y$ as vertices. This means that the set of medians of three points is simply the intersection of the rectangles.