Example 10 Let the set $A=\{1,2, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies 17 | $(a+$ $b)$, then $B$ is said to have property $P$.
(1) Find the number of binary subsets of $A$ that have property $P$;
(2) Find the number of a set of binary subsets of $A$, which are pairwise disjoint and have property $P$. (1994, Hebei Province High School Mathematics Competition)
Solution: (1) Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: $[0],[1], \cdots,[16]$.

Since $366=17 \times 21+9$, the classes $[1],[2], \cdots$, [9] each contain 22 numbers; [10], [11], $\cdots$, [16] and [0] each contain 21 numbers.
(i) When $a, b \in[0]$, the number of subsets with property $P$ is $\mathrm{C}_{21}^{2}=210$;
(ii) When $a \in[k], b \in[17-k], k=1,2$, $\cdots, 7$, the number of subsets with property $P$ is $\mathrm{C}_{22}^{1} \cdot \mathrm{C}_{21}^{1}=$ 462;
(iii) When $a \in[8], b \in[9]$, the number of subsets with property $P$ is $\mathrm{C}_{22}^{1} \cdot \mathrm{C}_{22}^{1}=484$.

Therefore, the number of subsets of $A$ with property $P$ is $210+462 \times 7+484=3928$.
(2) To ensure that the binary subsets do not intersect, when $a, b \in$ [0], 10 subsets can be paired;

When $a \in[k], b \in[17-k], k=1,2, \cdots, 7$, 21 subsets can be paired for each;

When $a \in[8], b \in$ [9], 22 subsets can be paired.

Therefore, the number of pairwise disjoint subsets with property $P$ is $10+21 \times 7+22=179$.

Note: Finding an appropriate criterion, i.e., using the remainder to partition the set, is the key to solving this problem.