To prove that for all $n > 1000$, we can arrange the numbers $1, 2, \dots, \binom{n}{2}$ on the edges of a complete graph with $n$ vertices so that the sum of the numbers assigned to the edges of any length three path (possibly closed) is not less than $3n - 1000 \log_2 \log_2 n$, we can proceed as follows:

1. Split the vertices into two sets:
Let $A$ and $B$ be two sets such that $0 \leq |A| - |B| \leq 1$. Specifically, let $A = \lceil \frac{n}{2} \rceil$ and $B = \lfloor \frac{n}{2} \rfloor$. This ensures that the sizes of $A$ and $B$ are as balanced as possible.

2. Assign numbers to edges between sets:
For any edge $e = xy$ where $x \in A$ and $y \in B$, assign an integer less than or equal to $|A||B| = \lfloor \frac{n^2}{4} \rfloor$. This means that the edges between the two sets $A$ and $B$ will have the smallest numbers assigned to them.

3. Assign remaining numbers to other edges:
Assign any of the remaining values to all the other edges, i.e., the edges within set $A$ and within set $B$.

4. Analyze the sum of numbers in any triangle:
Consider any triangle in the graph. At least one of the edges of the triangle must join two vertices that are either both in $A$ or both in $B$. The number assigned to this edge is at least $|A||B| + 1 = \lfloor \frac{n^2}{4} \rfloor + 1$.

5. Compare with the required bound:
Since $\lfloor \frac{n^2}{4} \rfloor + 1$ is a lower bound for the sum of the numbers assigned to the edges of any triangle, we need to show that this is greater than or equal to $3n - 1000 \log_2 \log_2 n$.

6. Verify the inequality:
$$\lfloor \frac{n^2}{4} \rfloor + 1 > 3n - 1000 \log_2 \log_2 n$$
For large $n$, $\frac{n^2}{4}$ grows much faster than $3n$ and $1000 \log_2 \log_2 n$. Therefore, for $n > 1000$, the inequality holds true.

$\blacksquare$