A sufficient (additional) condition is that $B$ *be compact for some Hausdorff vector topology for* $V$. The proof goes as follows.

Letting $\langle x_i\rangle$ be a Cauchy sequence, it is contained is some $nB$, and so it has some cluster point $y$ there. Given $\varepsilon>0$, there is $i_0$ such that $x_i-x_j\in\frac 12\varepsilon B$ for $i,j\ge i_0$ , and it remains to show that $x_{i_0}-y\in\frac 12\varepsilon B$ . Indeed, if this does not hold, we have $y\not\in x_{i_0}-\frac 12\varepsilon B$ . Since $B$ is compact, it is closed in the Hausdorff case, and so is $x_{i_0}-\frac 12\varepsilon B$ as we have a vector topology. By the cluster point property, there must be some $i\ge i_0$ with $x_i\not\in x_{i_0}-\frac 12\varepsilon B$ , which is impossible.