At most $n+1$ vectors, the angle between which $>\pi/2$. In a $n$ dimensional Euclidean space $V$, there exists at most $n+1$ vectors, each pair has inner product $<0$. 
This is geometrically obvious in $3$ dimensions...But how can we prove it analytically?

(Abridged from Robin Chapman's answer in a MO thread.)
You can have $m=n+1$. Take the vertices of a regular simplex with centre at the origin.
You can't have $m=n+2$. There is at least a two-dimensional space of vectors $(a_1,\ldots,a_{n+2})$ such that $$\sum_{i=1}^{n+2} a_i v_i=0.$$ This gives enough room for manoeuvre to ensure some $a_i>0$ and some $a_j<0$. Thus we get some nontrivial relation $$\sum_{i\in I}a_i v_i=\sum_{j\in J}b_j v_j\qquad\qquad(*)$$ where all the $a_i>0$ and $b_j>0$ and $I$ and $J$ are disjoint non-empty sets of indices. It follows that the dot product of the two sides of $(*)$ is negative, but that contradicts it being the square of the length of the left side.

(Abridged from Benoît Kloeckner's answer in the same MO thread.)
Let us prove by induction that this number is $n+1$. The result is obvious for $n=1$. Assume it for some $n$ and consider a set of mutually negative dot product vectors $v_0,v_1,\ldots, v_k$ in $\mathbb{R}^{n+1}$. Then all of $v_1,\ldots,v_k$ lie in the open half-space $\{\,v\mid v_0\cdot v<0\,\}$.
Now the orthogonal projections $v_i'$ of $v_i$ ($1\leqslant i\leqslant k$) on the hyperplane $\{\,v\mid v_0\cdot v=0\,\}$ satisfy $v_i'\cdot v_j'\leqslant v_i\cdot v_j$ by a direct computation (assuming $v_i$ are unitary, one has $v_i'=v_i-(v_i\cdot v_0)v_0$ so that $v_i'\cdot v_j' = v_i\cdot v_j-(v_i\cdot v_0)(v_j\cdot v_0)$). By induction $k$ is at most $n+1$ and we are done.
