C43 (36-6, Poland) Let $p$ be an odd prime, and consider the set $\{1,2, \cdots, 2 p\}$ of subsets $A$ that satisfy the following two conditions:
（1） $A$ has exactly $p$ elements;
（2） The sum of all elements in $A$ is divisible by $p$.

Find the number of all such subsets $A$.
Let $U=\{1,2, \cdots, p\}, V=\{p+1, p+2, \cdots, 2 p\}$, then $U, V$ are two subsets that satisfy the problem.

Let $W=\{1,2, \cdots, 2 p\}$, then $W$ has $\mathrm{C}_{2 p}^{p}$ $p$-element subsets. Excluding sets $U, V$, every $p$-element subset of $W$ has a non-empty intersection with $U, V$. Classify these subsets: Let $S, T$ be two $p$-element subsets of $W$ other than $U, V$, and satisfy
(1) $S \cap V=T \cap V$;
(2) If properly numbered, the elements $s_{1}, s_{2}, \cdots, s_{m}$ of $S \cap U$ and the elements $t_{1}, t_{2}, \cdots, t_{m}$ of $T \cap U$ satisfy the congruence system $s_{i}-t_{i} \equiv k$ $(\bmod p), i=1,2, \cdots, m$, for some $k \in\{0,1,2, \cdots, p-1\}$, then $S$ and $T$ are classified into the same category.

For different $k_{1}, k_{2}$ in the set $\{0,1,2, \cdots, p-1\}$, the sets obtained from $S$ are $T_{1}$, $T_{2}$, and the difference in the sum of elements is
$$
\sum_{i=1}^{p} s_{i}-\sum_{i=1}^{p} t_{i} \equiv\left(k_{1}-k_{2}\right) m(\bmod p),
$$

where $m$ is the number of elements in $S \cap U$. Since $S \cap U \neq \varnothing$, we have $0<m<p$, thus $\left(k_{1}-k_{2}\right) m \neq 0(\bmod p)$, i.e., $T_{1} \neq T_{2}$. Therefore, each class contains exactly $p$ sets, and the sums of the elements of these sets are distinct modulo $p$, so each class has exactly one set that satisfies condition (2).

In summary, among the $\mathrm{C}_{2 p}^{p}$ $p$-element subsets of $W=\{1,2, \cdots, 2 p\}$, excluding $U$ and $V$, every $p$ subsets are grouped into one class, and each class has exactly one subset that satisfies (2). Therefore, the total number of subsets of $W=\{1, 2, \cdots, 2 p\}$ that satisfy conditions (1) and (2) is
$$
\frac{1}{p}\left(C_{2 p}^{p}-2\right)+2 .
$$