Let $p$ be an odd prime number. We are tasked with finding the number of $p$-element subsets $A$ of the set $\{1, 2, \dots, 2p\}$ such that the sum of the elements in $A$ is divisible by $p$.

### Step 1: Representation of Subsets

The set $\{1, 2, \dots, 2p\}$ contains $2p$ elements. We want to choose $p$ elements from these $2p$ elements. The number of ways to choose $p$ elements from $2p$ is given by $\binom{2p}{p}$.

### Step 2: Applying Properties of Subsets and Congruences

When we consider a subset $A$ with $p$ elements, the sum of its elements, denoted as $\sum_{a \in A} a$, needs to satisfy:

$$\sum_{a \in A} a \equiv 0 \pmod{p}.$$

By properties of combinatorial numbers, when subsets are considered modulo a prime, symmetry and cancellation often play roles. Specifically, we will use the properties of binomial coefficients and modulo arithmetic.

### Step 3: Symmetry and Combinatorial Argument

Given the symmetry and periodicity modulo $p$, every element's contribution modulo $p$ can be considered as additive symmetric pairs $(x, 2p+1-x)$ whose sums modulo $p$ cover the full range as the whole set is considered.

### Step 4: Exploiting Known Theorems

Using Lucas' theorem or known properties about binomial coefficients modulo an odd prime $p$, it can be derived that:

$$\binom{2p}{p} \equiv 2 \pmod{p}.$$

This relation outlines that, up to a multiple of $p$, there are:

$$\binom{2p}{p} = 2p \times k + 2$$

possible selections of sets, where the residue class modulo $p$, i.e., remainder when divided by $p$, is $2$.

### Conclusion

Since we are dividing the total number of such subsets by $p$ to get those subsets where the sum is specifically zero modulo $p$, the formula for the number of such subsets is:

$$\frac{1}{p} \left( \binom{2p}{p} - 2 \right) + 2.$$

This simplifies to:

$$\boxed{2 + \frac{1}{p} \left( \binom{2p}{p} - 2 \right)}.$$

Thus, the number of $p$-element subsets $A$ such that the sum of the elements is divisible by $p$ is given by the expression inside the boxed formula.