The 54th question: Let integer $n \geq 2$, if for a given set of $n$ points in the plane with no three points collinear, there exist $m$ lines such that no line passes through any of the given points; for any two distinct points $X, Y$ chosen from the $n$ points, there exists a line such that $X, Y$ are on opposite sides of the line. Find the smallest positive integer $m$.

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Question 54:
Solution: First, take points $\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ clockwise on the circumference (subscripts are taken modulo $\mathrm{n}$), then the $\mathrm{n}$ chords $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{i}+1}(1 \leq \mathrm{i} \leq \mathrm{n})$ all intersect with at least one line. Since each line intersects with at most two chords, at least $\left[\frac{n}{2}\right]$ lines are needed.
Next, we prove that $\left[\frac{n}{2}\right]$ lines satisfy the requirement.
First, we prove a lemma: For any four distinct points $P_{1}, P_{2}, Q_{1}, Q_{2}$, there always exists a line that separates $P_{1}, P_{2}$ on one side and $Q_{1}, Q_{2}$ on the other side, and this line does not pass through any other given points.

Proof of the lemma: In fact, take any point $\mathrm{X}$ on the segment $\mathrm{P}_{1} \mathrm{P}_{2}$ and any point $\mathrm{Y}$ on the segment $\mathrm{Q}_{1} \mathrm{Q}_{2}$, such that the line $X Y$ does not coincide with the lines $P_{1} P_{2}$ and $Q_{1} Q_{2}$, and does not pass through any other points. Such a line $X Y$ clearly exists and there are infinitely many such lines. These lines all satisfy the condition. The lemma is proved.
Now, let's return to the original problem.
(1) First, consider the case where $\mathrm{n}$ is even.
Determine a line that is not parallel to any line connecting any two points (since $\mathrm{n}$ is finite, this is achievable). Then, translate this line in the plane so that it sweeps through the given points one by one. When this line has no given points and there are $\frac{n}{2}$ points on each side, we call it the first line. This divides the plane into two regions, which we can call the left and the right.

Then, add lines in the following manner: If there are still unseparated point pairs on both the left and the right, then according to the lemma, we can add a line to separate these two point pairs; If only one side has unseparated point pairs, then select a point pair on this side and add a line to separate them.

Following this method, after adding $\frac{n}{2}-1$ lines, there will be at least $\frac{n}{2}$ regions on the left (or right), with each region containing at most one given point. Therefore, these lines satisfy the condition.
(2) If $\mathrm{n}$ is odd. First, take any point $\mathrm{A}$, remove point $\mathrm{A}$, and then for the remaining $\mathrm{n}-1$ points, select $\frac{\mathrm{n}-1}{2}$ lines according to (1) so that each point is in a separate region.

After adding back point $\mathrm{A}$, if $\mathrm{A}$ is not in the same region as any other point, then these $\frac{\mathrm{n}-1}{2}$ lines satisfy the condition; If $\mathrm{A}$ is in the same region as some point $\mathrm{B}$, then add a line to separate $\mathrm{A}$ and $\mathrm{B}$. Such $\frac{\mathrm{n}-1}{2}+1=\left\lceil\frac{n}{2}\right]$ lines satisfy the requirement. In summary, the minimum value of $\mathrm{m}$ is $\left\lceil\left.\frac{\mathrm{n}}{2} \right\rvert\,\right.$.