Let $p$ be an odd prime number. How many $p$-element subsets $A$ of $\{1,2,\ldots \ 2p\}$ are there, the sum of whose elements is divisible by $p$?
1. **Define the sets and their properties:**
   Let \( X = \{1, 2, \ldots, p\} \) and \( Y = \{p+1, p+2, \ldots, 2p\} \). Note that \( X \) and \( Y \) form a complete residue system modulo \( p \). This means that every integer from \( 1 \) to \( 2p \) can be represented uniquely modulo \( p \) by an element in \( X \cup Y \).

2. **Translation and trivial subsets:**
   For a subset \( A \subset X \) and \( z \in \mathbb{Z}/p\mathbb{Z} \), define \( A + z = \{a + z \mod p \mid a \in A\} \). Similarly, for a subset \( B \subset Y \), define \( B + z = \{b + z \mod p \mid b \in B\} \).

   A subset \( A \subset X \) is called trivial if \( A = \emptyset \) or \( A = X \). Similarly, a subset \( B \subset Y \) is trivial if \( B = \emptyset \) or \( B = Y \).

3. **Nontrivial subsets and translation:**
   Consider a nontrivial subset \( P \subset (X \cup Y) \). Define a translation for \( P \):
   - If \( P \cap X \) is nontrivial, define \( P + z = ((P \cap X) + z) \cup (P \cap Y) \).
   - If \( P \cap Y \) is nontrivial, define \( P + z = (P \cap X) \cup ((P \cap Y) + z) \).

4. **Sum of elements modulo \( p \):**
   When \( z \) goes through all possible residue classes modulo \( p \), the sum of the elements of \( P + z \) also goes through all possible residue classes modulo \( p \). Therefore, there is exactly one value of \( z \) for which the sum of the elements of \( P + z \) is divisible by \( p \).

5. **Counting the subsets:**
   There are exactly \( \binom{2p}{p} \) subsets of \( \{1, 2, \ldots, 2p\} \) with \( p \) elements. Among these, 2 subsets are trivial (the subsets \( X \) and \( Y \)). The nontrivial subsets can be divided into equivalence classes of \( p \) sets each, where each class has exactly one subset whose sum of elements is divisible by \( p \).

6. **Final calculation:**
   The number of nontrivial \( p \)-element subsets is \( \binom{2p}{p} - 2 \). Since each equivalence class contains \( p \) subsets, the number of equivalence classes is \( \frac{\binom{2p}{p} - 2}{p} \). Adding the 2 trivial subsets, the total number of \( p \)-element subsets whose sum is divisible by \( p \) is:
   \[
   \frac{\binom{2p}{p} - 2}{p} + 2
   \]

The final answer is \(\boxed{\frac{\binom{2p}{p} - 2}{p} + 2}\)