Monica and Bogdan are playing a game, depending on given integers $n, k$. First, Monica writes some $k$ positive numbers. Bogdan wins, if he is able to find $n$ points on the plane with the following property: for any number $m$ written by Monica, there are some two points chosen by Bogdan with distance exactly $m$ between them. Otherwise, Monica wins.

Determine who has a winning strategy depending on $n, k$.

[i](Proposed by Fedir Yudin)[/i]
1. **Claim**: Monica wins if and only if \( n \leq k \).

2. **Case 1**: \( n \geq k+1 \)
   - Bogdan can pick the origin \((0,0)\) and the points \((x_i, 0)\) for \(1 \leq i \leq k\), where the \(x_i\) are the specified distances.
   - This ensures that for each \(m\) written by Monica, there exists a pair of points \((0,0)\) and \((x_i, 0)\) such that the distance between them is exactly \(m\).
   - Therefore, Bogdan wins in this case.

3. **Case 2**: \( n \leq k \)
   - Monica can pick the distances \(1, 2, 2^2, \ldots, 2^{k-1}\).
   - Assume for contradiction that Bogdan could win. Consider the graph \(G\) with vertices being the points picked by Bogdan and edges between vertices with distances belonging to those specified by Monica.
   - Since the distances are \(1, 2, 2^2, \ldots, 2^{k-1}\), all edge lengths are distinct.
   - There can't be any cycle in \(G\) because \(2^t > 2^{t-1} + \cdots + 1\) for all \(t\). This implies that \(G\) is a forest (a union of trees).
   - For each tree in the forest, the number of edges \(E_i\) and vertices \(V_i\) satisfy \(E_i = V_i - 1\).
   - Summing over all trees, we get \(E = V - T\), where \(T\) is the number of trees.
   - Since \(V = n\) and \(E \geq k\), we have \(n \geq k + T\).
   - Given \(n \leq k\), this implies \(T = 1\), so \(n = k + 1\), which contradicts \(n \leq k\).
   - Therefore, Monica wins in this case.

The final answer is Monica wins if and only if \( \boxed{ n \leq k } \).