Let $k$ be a positive integer and $P$ a point in the plane. We wish to draw lines, none passing through $P$, in such a way that any ray starting from $P$ intersects at least $k$ of these lines. Determine the smallest number of lines needed.
1. **Construction of the Lines:**
   - Consider a regular $(2k+1)$-gon $V_1V_2...V_{2k+1}$ centered at point $P$.
   - Draw the $2k+1$ lines $V_iV_{i+1}$ for $i=1,2,...,2k+1$ (indices taken modulo $2k+1$).
   - When these lines are extended, they divide the plane into $2k+1$ sectors, each containing $P$ in its interior.
   - Any ray starting from $P$ will intersect the sides of these sectors. Since there are $2k+1$ sectors and each sector is bounded by two lines, any ray from $P$ will intersect at least $k$ of these lines.

2. **Proof of Necessity:**
   - Suppose it is possible to achieve the goal with only $2k$ lines.
   - Consider a line $\ell$ passing through $P$ and parallel to one of the $2k$ lines. This line $\ell$ divides the plane into two rays.
   - Since $\ell$ is parallel to one of the $2k$ lines, it intersects the remaining $2k-1$ lines at most $2k-1$ times.
   - Therefore, there exists a ray starting from $P$ that intersects at most $2k-1$ lines.
   - This implies that there is a ray starting from $P$ that intersects fewer than $k$ lines, which contradicts the requirement that any ray starting from $P$ must intersect at least $k$ lines.

Thus, the smallest number of lines needed is $2k+1$.

The final answer is $\boxed{2k+1}$