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 $a_1,a_2,...,a_n$ such that $a_1 + a_2 + ... + a_n = S$ and $a_n-a_1=d.$
To find all pairs of positive integers \((n, d)\) such that for every integer \(S\), there exists a unique non-decreasing sequence of \(n\) integers \(a_1, a_2, \ldots, a_n\) satisfying \(a_1 + a_2 + \cdots + a_n = S\) and \(a_n - a_1 = d\), we will analyze different cases based on the value of \(n\).

### Case 1: \(n = 1\)
If \(n = 1\), we have only one integer \(a_1\). The conditions become:
\[ a_1 = S \]
\[ a_1 - a_1 = 0 \]
Thus, \(d = 0\). Therefore, the pair \((1, 0)\) satisfies the conditions.

### Case 2: \(n = 2\)
If \(n = 2\), we have two integers \(a_1\) and \(a_2\). The conditions become:
\[ a_1 + a_2 = S \]
\[ a_2 - a_1 = d \]
Adding these equations, we get:
\[ 2a_2 = S + d \]
\[ a_2 = \frac{S + d}{2} \]
Subtracting the second equation from the first, we get:
\[ 2a_1 = S - d \]
\[ a_1 = \frac{S - d}{2} \]
For \(a_1\) and \(a_2\) to be integers, \(S + d\) and \(S - d\) must both be even, implying that \(S\) and \(d\) must have the same parity. However, this does not guarantee uniqueness for all \(S\). Therefore, there are no solutions for \(n = 2\).

### Case 3: \(n = 3\)
If \(n = 3\), we have three integers \(a_1, a_2, a_3\). The conditions become:
\[ a_1 + a_2 + a_3 = S \]
\[ a_3 - a_1 = d \]
Let us define \(b_k = a_k - a_1\). We can rewrite the first condition as:
\[ 3a_1 + b_2 + b_3 + d = S \]
Taking this modulo 3, we get:
\[ b_2 + b_3 + d \equiv S \pmod{3} \]
Since \(b_2\) and \(b_3\) are non-negative and \(a_1 \leq a_2 \leq a_3\), we have \(0 \leq b_2 \leq b_3 \leq d\).

For uniqueness, we must ensure that the sequence can reach all values modulo 3. We analyze the possible values of \(d\):

- **If \(d = 1\)**: We get \(b_2 + b_3 + 1 \equiv S \pmod{3}\). Since \(b_2\) and \(b_3\) can be 0 or 1, we cannot achieve all values modulo 3. Thus, \(d = 1\) does not work.
- **If \(d = 2\)**: We get \(b_2 + b_3 + 2 \equiv S \pmod{3}\). Since \(b_2\) and \(b_3\) can be 0, 1, or 2, we can achieve all values modulo 3 uniquely. Thus, \(d = 2\) works.
- **If \(d = 3\)**: We get \(b_2 + b_3 + 3 \equiv S \pmod{3}\). Since \(b_2\) and \(b_3\) can be 0, 1, 2, or 3, we can achieve all values modulo 3, but not uniquely. Thus, \(d = 3\) does not work.

Therefore, the pair \((3, 2)\) satisfies the conditions.

### Case 4: \(n \ge 4\)
For \(n \ge 4\), we need to check if the conditions can be satisfied uniquely. Consider \(d = 2\):
- We can represent the number 4 as \(a_1 + a_2 + \cdots + a_n\) in multiple ways, breaking the uniqueness requirement. For example:
  - \(a_i = 0\) for all \(i < n-3\) and \(a_i = 1\) for all \(n-3 \le i \le n\).
  - \(a_i = 0\) for all \(1 \le i \le n-2\) and \(a_i = 2\) for all \(n-1 \le i \le n\).

Similarly, for \(d = 3\), we can represent the number 6 in multiple ways, breaking the uniqueness requirement. For example:
- \(a_i = 0\) for all \(1 \le i \le n-2\) and \(a_i = 3\) for all \(n-1 \le i \le n\).
- \(a_i = 0\) for all \(1 \le i \le n-3\) and \(a_{n-2} = 1\), \(a_{n-1} = 2\), and \(a_n = 3\).

Therefore, there are no solutions for \(n \ge 4\).

Hence, the only solutions are \(\boxed{(n, d) = \{(1, 0), (3, 2)\}}\).