Problem 1311

1. Define the problem in terms of "good" integers:
We say that a positive integer $n$ is good if there exist positive integers $a_1 < a_2 < \cdots < a_k$ such that the sums $a_i + a_j$ for $1 \leq i < j \leq k$ are pairwise distinct, and among those sums, there are $n$ consecutive integers.

2. Base case:
Clearly, $1$ is good. For instance, if we take $a_1 = 1$ and $a_2 = 2$, then the sum $a_1 + a_2 = 3$ is a single sum, which is trivially consecutive.

3. Inductive step:
Suppose $n$ is good. This means there exist positive integers $a_1 < a_2 < \cdots < a_k$ such that the sums $a_i + a_j$ for $1 \leq i < j \leq k$ include $n$ consecutive integers, say $t+1, t+2, \ldots, t+n$.

4. **Case 1: $t+n+1$ is among the sums:**
If $t+n+1$ is already among the sums, then we have $n+1$ consecutive integers, and thus $n+1$ is good.

5. **Case 2: $t+n+1$ is not among the sums:**
If $t+n+1$ is not among the sums, we need to construct a new set of integers $a_1', a_2', \ldots, a_{k+1}'$ such that the sums include $t+1, t+2, \ldots, t+n+1$.

Consider the integers:
$$t+n+1 - a_{k+1} < a_1 < a_2 < \cdots < a_k < a_{k+1}$$
where $a_{k+1}$ is a sufficiently large positive integer that ensures all sums are pairwise distinct.

6. **Ensuring $t+n+1 - a_{k+1}$ is positive:**
To ensure $t+n+1 - a_{k+1}$ is positive, we can add a large integer $M$ to all $a_i$. This does not affect the pairwise distinctness of the sums but shifts all sums by $2M$. Thus, we can always choose $M$ such that $t+n+1 - a_{k+1} + M > 0$.

7. Conclusion:
By the inductive step, if $n$ is good, then $n+1$ is good. Since $1$ is good, by induction, every positive integer is good. Therefore, $1000$ is good.

$\blacksquare$