For a positive integer $n$, let $\mathrm{I}_{\mathrm{n}}=\{1,2, \ldots, \mathrm{n}\}, \mathrm{P}_{\mathrm{n}}=\left\{\left.\frac{\mathrm{m}}{\sqrt{k}} \right\rvert\, \mathrm{m} \in \mathrm{I}_{\mathrm{n}}, \mathrm{k} \in \mathrm{I}_{\mathrm{n}}\right\}$. If the sum of any two elements in a subset $\mathrm{A}$ of $\mathrm{P}_{\mathrm{n}}$ is not a perfect square, then $\mathrm{A}$ is called a “sparse set”. Find the maximum value of $\mathrm{n}$ such that $\mathrm{P}_{\mathrm{n}}$ can be partitioned into the union of two disjoint sparse sets.
Question 229, Solution: We first prove that $n \leq 14$. Otherwise, if $n \geq 15$, then $\{1,2, \ldots, 15\} \subset P_{n}$. If $P_{n}$ can be divided into the union of two disjoint sparse sets $A$ and $B$, without loss of generality, let $1 \in A$, then $3 \in B$, thus $6 \in A$, then $10 \in B$, and consequently $15 \in A$. However, at this point $1 \in A, 15 \in A, 1+15=4^{2}$, which contradicts the fact that $A$ is a sparse set. Therefore, $n \leq 14$.

On the other hand, we can divide $P_{14}$ into the union of two disjoint sparse sets. Let $A=\{1,2, \ldots, 14\}$, $B=\left\{\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \ldots, \frac{13}{2}\right\}$, $C=\left\{\frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{7}{3}, \frac{8}{3}, \frac{10}{3}, \frac{11}{3}, \frac{13}{3}, \frac{14}{3}\right\}$, and $D=\left\{\left.\frac{m}{\sqrt{k}} \right| m \in I_{14}, k \in I_{14}, \sqrt{k} \notin Z\right\}$, then $P_{14}=A \cup B \cup C \cup D$. Note that the elements in set $D$ are all irrational numbers, and adding any of them to any other element in $P_{n}$ cannot result in a perfect square, so we only need to consider dividing sets $A$, $B$, and $C$ into sparse sets. Let
$M_{1}=\{1,2,4,6,9,11,13\}, N_{1}=\{3,5,7,8,10,12,14\};$
$M_{2}=\left\{\frac{1}{2}, \frac{5}{2}, \frac{9}{2}, \frac{11}{2}\right\}, N_{2}=\left\{\frac{3}{2}, \frac{7}{2}, \frac{13}{2}\right\}$
$M_{3}=\left\{\frac{1}{3}, \frac{4}{3}, \frac{5}{3}, \frac{10}{3}, \frac{13}{3}\right\}, N_{3}=\left\{\frac{2}{3}, \frac{7}{3}, \frac{8}{3}, \frac{11}{3}, \frac{14}{3}\right\}$.
Let $M=M_{1} \cup M_{2} \cup M_{3} \cup D, N=N_{1} \cup N_{2} \cup N_{3}$, then $M$ and $N$ are both sparse sets, and $M \cap N=\Phi$, $M \cup N=P_{14}$.
In conclusion, the maximum value of $n$ that satisfies the condition is 14.