Find, with proof, all pairs of positive integers $(n, d)$ with the following property: for every integer $S$, there exists a unique non-decreasing sequence of n integers $a_1$, $a_2$, $\cdots$, $a_n$ such that $a_1 + a_2 + \cdots + a_n = S$ and $a_n-a_1 = d$.
Let $d_i = a_i - a_1$, then the sequence can be written as $$\underbrace{a_1, a_1 + d_2, a_1+d_3,\cdots, a_1+d_{n-1}, a_1 + d}_{n}$$
where $$d_2, d_3, \cdots, d_{n-1}$$ forms a non-decreasing sequence whose elements are non-negative integers and not exceeding $d$. By the conclusion of # 4145, there are totally $C_{n-2+d}^d$ such sequences.
We claim that in order to meet the requirement, the number of difference sequence must equal $n$. This is because the sum of this sequence is $$\begin{array}{rl}&S=n a_1 + d + d_2+d_3+\cdots + d_{n-1}\\ \implies& S\equiv d+d_2+d_3+\cdots +d_{n-1}\pmod{n}\end{array}$$
The right side of the above relation must enumerate every value of $0$, $1$, $\cdots$, $n-1$, once and only once. Now let's solve $$C_{n-2+d}^d=n$$ by analyzing the value of $d$.
When $d=1$, the relation cannot hold.
When $d=2$, we find $n=3$ can satisfy the relation.
When $d=3$, there is no integer solution.
When $d\ge 4$, the equation can be simplified to $(n+2)(n+1)(n-1)=2\times 3\times 4$. Clearly, $n=1,2$ are not solution. When $n\ge 3$, we have $(n+2) > 4$, $(n+1) > 3$, $(n-1)\ge 2$.
Therefore, the only possible solutions $(n, d) = \boxed{(3, 2)}$. And in dead, it is a solution. This is because in this case, there are three sequences:
$$\begin{array}{rcl} a_1, a_1, a_1 + 2 & \implies & S= 3a_1 + 2 \\ a_1, a_1 + 1, a_1 + 2 & \implies & S= 3a_1 + 3 \\ a_1, a_1 +2, a_1 +2 &\implies & S= 3a_1 + 4 \end{array}$$
Hence, every integer $S$ can be uniquely represented by one of these three sequences, depending on the value of $(S \mod{3})$.