Claim: Monica wins if and only if $n\leq k$. If $n\geq k+1$, then Bogdan picks the origin and the points $(x_i,0)$ ($1\leq i\leq k$) where the $x_i$ are the specified distances. If $n\leq k$ then it suffices for Monica to pick the distances $1,2,2^2,...,2^{k-1}$. Assume Bogdan could win. Consider the graph with the vertices picked by Bogdan and the edges are between vertices with distances belonging to those specified by Monica (if a certain distance appears more than once, pick only one example at random). There can't be any cycle, because $2^t>2^{t-1}+\cdots +1$ for all $t$, and all edge lengths are distinct. Therefore the graph is a union of trees. Each of them satisfies $E_i+1=V_i$, which means that in the total of the graph we have $n=V\geq E+1\geq k+1$. However this is a contradiction with $n\leq k$, so we are done.