Solution:

For each triangle $\mathcal{T}$, we let $p(\mathcal{T})$ denote the perimeter of $\mathcal{T}$.

First, we claim that $\frac{1}{2} p(\triangle ABE) < p(\triangle ADE) < 2 p(\triangle ABE)$. To see why, observe that
$$p(\triangle ADE) = EA + ED + AD < 2(EA + ED) = 2(EA + EB) < 2 p(\triangle ABE)$$
Similarly, one can show that $p(\triangle ABE) < 2 p(\triangle ADE)$, proving the desired inequality.

This inequality limits the possibility of $n$ to only those in $(5\pi, 20\pi) \subset (15.7, 62.9)$, so $n$ could only range from $16, 17, 18, \ldots, 62$, giving $47$ values. These values are all achievable because

- when $AD$ approaches zero, we have $p(\triangle ADE) \rightarrow 2EA$ and $p(\triangle ABE) \rightarrow 4EA$, implying that $p(\triangle ADE) \rightarrow \frac{1}{2} p(\triangle ABE) = 5\pi$;
- similarly, when $AB$ approaches zero, we have $p(\triangle ADE) \rightarrow 2 p(\triangle ABE) = 20\pi$; and
- by continuously rotating segments $AC$ and $BD$ about $E$, we have that $p(\triangle ADE)$ can reach any value between $(5\pi, 20\pi)$.

Hence, the answer is $47$.