Question 222: Divide the set $S=\{1,2, \ldots, 36\}$ into $k$ mutually disjoint non-empty subsets $A_{1}, A_{2}, \ldots, A_{k}$ whose union is $S$. If for each $A_{i}(1 \leq i \leq k)$, the sum of any two distinct elements is not a perfect square, find the minimum possible value of $k$.
Question 222, Solution Method One: Solve the system of equations $\left\{\begin{array}{l}a+b=49 \\ b+c=36 \\ c+a=25\end{array}\right.$, to get $\left\{\begin{array}{l}a=19 \\ b=30 \\ c=6\end{array}\right.$, thus we know that $6, 19, 30$ are not in the same set, hence $k \geq 3$.
$$
\begin{array}{l}
\text { In one way, when } k=3 \text {, we have } \\
A_{1}=\{4 k+3 \mid 0 \leq k \leq 8\} \cup\{4,8,16,24,36\} ; \\
A_{2}=\{4 k+1 \mid 0 \leq k \leq 8\} \cup\{6,14,18,26,34\} ; \\
A_{3}=\{2,10,12,20,22,28,30,30,32\}
\end{array}
$$

which satisfies the conditions.
In summary, the smallest value of $k$ that satisfies the conditions is 3.
Solution Method Two: Clearly, $k>1$. If $k=2$, assume $1 \in A_{1}$, then $3 \in A_{2}$, and consequently $6 \in A_{1}, 10 \in A_{2}$, $15 \in A_{1}$. But $1+15=4^{2}$, which leads to a contradiction. Hence, $k \geq 3$.
$$
\begin{array}{l}
\text { In another way, when } k=3 \text {, we take } \\
A_{1}=\{4 k+3 \mid 0 \leq k \leq 8\} \cup\{4,8,16,24,36\} ; \\
A_{2}=\{4 k+1 \mid 0 \leq k \leq 8\} \cup\{6,14,18,26,34\} ; \\
A_{3}=\{2,10,12,20,22,28,30,32\}
\end{array}
$$

which satisfies the conditions.
In summary, the smallest value of $k$ that satisfies the conditions is 3.