1. Understanding the Problem:
- We are given a rectangle $ABCD$ with diagonals intersecting at $E$.
- The perimeter of triangle $ABE$ is $10\pi$.
- We need to find the number of possible integer values for the perimeter of triangle $ADE$, denoted as $n$.

2. Setting Up the Equations:
- Let $a$ and $b$ be the lengths of the sides $AB$ and $AD$ of the rectangle, respectively.
- The diagonals of the rectangle are equal, and they intersect at $E$, the midpoint of each diagonal.
- The length of each diagonal is $\sqrt{a^2 + b^2}$.

3. **Perimeter of Triangle $ABE$:**
- The perimeter of $\triangle ABE$ is given by:
$$a + \sqrt{a^2 + b^2} + \frac{\sqrt{a^2 + b^2}}{2} = 10\pi$$
- Simplifying, we get:
$$a + \frac{3\sqrt{a^2 + b^2}}{2} = 10\pi$$

4. **Perimeter of Triangle $ADE$:**
- The perimeter of $\triangle ADE$ is given by:
$$b + \sqrt{a^2 + b^2} + \frac{\sqrt{a^2 + b^2}}{2} = n$$
- Simplifying, we get:
$$b + \frac{3\sqrt{a^2 + b^2}}{2} = n$$

5. **Analyzing the Range of $n$:**
- As $a \rightarrow 0$, $b \rightarrow 10\pi$, and thus:
$$n \rightarrow 20\pi$$
- As $b \rightarrow 0$, $a \rightarrow 10\pi$, and thus:
$$n \rightarrow 5\pi$$

6. Finding the Integer Values:
- The range of $n$ is from $5\pi$ to $20\pi$.
- Converting to integer values:
$$\lfloor 20\pi \rfloor - \lceil 5\pi \rceil + 1$$
- Approximating $\pi \approx 3.14$:
$$\lfloor 20 \times 3.14 \rfloor = \lfloor 62.8 \rfloor = 62$$
$$\lceil 5 \times 3.14 \rceil = \lceil 15.7 \rceil = 16$$
- Therefore, the number of integer values is:
$$62 - 16 + 1 = 47$$

The final answer is $\boxed{47}$.