1.28 Let the set $A=\{1,2,3, \cdots, 366\}$. If a two-element subset $B=\{a, b\}$ of $A$ satisfies $17 \mid (a+b)$, then $B$ is said to have property $P$.
(1) Find the number of two-element subsets of $A$ that have property $P$;
(2) Find the maximum number of pairwise disjoint two-element subsets of $A$ that have property $P$.
[Solution] $17 \mid(a+b)$ is equivalent to $a+b \equiv 0(\bmod 17)$, which holds if and only if
$$
\begin{aligned}
a \equiv k(\bmod 17), b \equiv 17-k(\bmod 17), k=0,1, \cdots, 16 . \\
\text { Let } \\
\left.A_{k}=|n| n \equiv k(\bmod 17), n \in A\right\}, k=0,1, \cdots, 16, \text { then } A \\
=\bigcup_{k=0}^{16} A_{k} \text { and }\left|A_{k}\right|=22, k=1,2, \cdots, 9 ;\left|A_{k}\right|=21, k=10,11, \cdots, 16,0 .
\end{aligned}
$$

Thus, (1) can be rewritten as: $17 \mid(a+b)$ if and only if
$$
a \in A_{k}, b \in A_{17-k}, k=0,1, \cdots, 16 \text {, }
$$

where $A_{17}=A_{0}$.
When $a, b \in A_{0}$, the number of subsets with property $P$ is $C_{21}^{2}=210$.
When $1 \leqslant k \leqslant 7, a \in A_{k}, b \in A_{17-k}$, the number of two-element subsets with property $P$ is $21 \times 22=462$.

When $a \in A_{8}, b \in A_{9}$, the number of two-element subsets with property $P$ is $22^{2}=$ 484.
In summary, the total number of two-element subsets with property $P$ is
$$
210+462 \times 7+484=3928 .
$$

To calculate (2), note that when $a \in A_{0}, b \in A_{0}$, since $\left|A_{0}\right|=21$, the maximum number of pairwise disjoint two-element subsets is 10. For $1 \leqslant k \leqslant 7, a \in A_{k}, b \in$ $A_{17-k}$. Since $\left|A_{k}\right|=22,\left|A_{17-k}\right|=21$, the maximum number of pairwise disjoint two-element subsets is 21. Similarly, when $a \in A_{8}, b \in A_{9}$, the maximum number of pairwise disjoint two-element subsets is 22. Therefore, the maximum number of pairwise disjoint two-element subsets with property $P$ is
$$
10+21 \times 7+22=179 \text {. }
$$