First Solution. Defining $b_i = a_{i+1} - a_i$, $i = 1, \dots, n-1$. Thus, $(b_i)$ is a decreasing sequence of positive real numbers. Then, the difference of any two elements of $A$ can be written as the sum of some of the $b_i$'s. Assume
$$b_{i_1} + b_{i_2} + \dots + b_{i_j} = b_{i_2} + \dots + b_{i_j} = \dots = b_{i_m} + \dots + b_{i_n} = 1$$
Such that $i_1 < i_2 < \cdots < i_m$. If $t > k$ then $j_t - i_t > j_k - i_k$. Since these numbers are positive, we find that $i_1, \dots, i_m$ are distinct. Therefore,
$$j_m = j_m - i_m + i_m \geq m - 1 + m$$
Yielding, $n \ge 2m$. Now, we construct two different examples;

1st Example. Let $b_m = \frac{2m-1}{m(3m-1)}$, $b_{m+1} = \frac{2m-2}{m(3m+1)}$, $\dots$, $b_{2m-1} = \frac{2}{m(3m-1)}$. Then, we go inductively through $i = m-1$ and construct $b_1, \dots, b_{m-1}$ as $b_i = b_{2i}+b_{2i+1}$, notice that since for all $i < n-1$ the inequalities $b_i > b_{i+1}$ and $b_{2i} + b_{2i+1} > b_{2i+2} + b_{2i+3}$ are equivalent, we find that $b_1, \dots$ form a decreasing sequence. On the other hand,
$$b_i + \cdots + b_{2i-1} = b_{i+1} + \cdots + b_{2(i+1)-1}$$
It is easy to verify that such an example works.

Second Solution We can prove something more general; for every positive integer $m$ there is a nice set $A$ of size $2m$ and a nonzero real number $d$ that can be written as a difference of two elements of $A$ in at least $m$ different ways. Let $S_d = \{(a,b) \in A \times A; a-b=d\}$, we prove that $|S_d| \ge m$. Indeed, Let $A = \{a_1 < \cdots < a_{2m}\}$, which is made of two halves. The set $A$ begins with $0$, and then has gaps $1+ (i-1)\delta$, for some very small $\delta > 0$ which will be specified later. The first half of $A$ is filled like this. That is, for $1 \le k \le m+1$, we define $a_k = (k-1)+\delta \frac{(k-2)(k-1)}{2}$, and so the first $m+1$ elements of $A$ are elements of the set $A_1 = \{0,1,2+\delta,3+3\delta,\cdots,m+\delta \frac{m(m-1)}{2}\}$. Fix $d = m + \delta \frac{m(m-1)}{2} = a_{m+1}$. The rest of $A$ is defined iteratively. For $1 \le i \le m-1$, we get $a_{m+1+i} = a_{1+2i}+d$. This immediately gives rise to the system of equations $d = a_{m+1} - a_1 = a_{m+2} - a_3 = \cdots = a_{2m} - a_{2m-1}$. We therefore $|S_d(A)| \ge m$. It remains to check that this set is nice. Note that $A_1 = \{a_1, \cdots, a_{m+1}\}$ is nice, since the consecutive difference increase by $\delta$ at each step.
We will prove by induction on $i$ that the set $\{a_1, \cdots, a_{m+2+i}\}$ is nice for $0 \le i \le m-2$. We first check the base for $i=0$. We need to verify that the difference $a_{m+2}-a_{m+1}$ is sufficiently large, which will give a condition on $\delta$. We must have $a_{m+2}-a_{m+1}>a_{m+1}-a_m$. We need to have $\delta < \frac{1}{m-2}$. Now, let $1 \le i \le m-2$. We must verify that $\{a_1, \cdots, a_{m+2+i}\}$ is nice, given the induction hypothesis that $\{a_1, a_2, \cdots, a_{m+1+i}\}$ is nice. All that remains to check that $a_{m+2+i}-a_{m+1+i} > a_{m+1+i}-a_{m+i}$. Note That $a_{m+2+i}-a_{m+1+i} = a_{1+2(i+1)} - a_{1+2i}$ and $a_{m+1+i} - a_{m+i} = a_{1+2i} - a_{1+2(i-1)}$. Note that, since the difference on the right hand side above are then consecutive differences of length two within a nice set, we have
$$a_{m+2+i} - a_{m+1+i} = a_{1+2(i+1)} - a_{1+2i} > a_{1+2i} - a_{1+2(i-1)} = a_{m+1+i} - a_{m+i}$$
as needed. Here we have used the inductive hypothesis that $\{a_1, \cdots, a_{m+1+i}\}$ is nice as well as the fact that $1+2(i+1) \le m+1+i$. The latter inequality follows from the condition $i \le m-2$.