USAJMO

To find all sequences $x_1, x_2, \ldots, x_{n-1}$ of positive integers satisfying the conditions given in the problem, we proceed as follows:

Given Conditions:
1. $x_1 < x_2 < \cdots < x_{n-1}$.
2. $x_i + x_{n-i} = 2n$ for all $i = 1, 2, \ldots, n-1$.
3. For any indices $i$ and $j$ where $x_i + x_j < 2n$, there exists an index $k$ such that $x_i + x_j = x_k$.

Objective: Find all sequences $x_1, x_2, \ldots, x_{n-1}$.

### Step-by-Step Solution:

Step 1: Understand the implications of conditions (a) and (b).

From condition (b), we have $x_i + x_{n-i} = 2n$. Let's express $x_i$ in terms of $i$.

Notice if $x_i = 2i$, then:
$$x_{n-i} = 2(n-i),$$
and,
$$x_i + x_{n-i} = 2i + 2(n-i) = 2n.$$
Thus, $x_i = 2i$ satisfies condition (b) for all $i = 1, 2, \ldots, n-1$.

**Step 2: Check condition (a): $x_1 < x_2 < \cdots < x_{n-1}$.**

For the proposed sequence $x_i = 2i$, observe:
$$x_1 = 2 \cdot 1 < x_2 = 2 \cdot 2 < \cdots < x_{n-1} = 2(n-1).$$
This clearly satisfies condition (a).

Step 3: Verify condition (c).

Given any two indices $i, j$, suppose that $x_i + x_j < 2n$.

For $x_i = 2i$ and $x_j = 2j$, the sum is:
$$x_i + x_j = 2i + 2j = 2(i + j).$$
If $2(i + j) < 2n$, then $i + j < n$.

For $k = i + j$, it follows that $x_k = 2(i + j) = x_i + x_j$.

Thus, for this arrangement, all three conditions are satisfied:
- The sequence is strictly increasing.
- The sum conditions are met.
- Every necessary sum of two indices corresponds directly to another term in the sequence.

Therefore, the sequence that satisfies all conditions is:
$$x_i = 2i \quad \text{for} \quad i = 1, 2, \ldots, n-1.$$

Final Answer:

The sequences that satisfy all the given properties are:
$$\boxed{x_i = 2i \text{ for } i = 1, 2, \ldots, n-1}.$$