Problem 1027

To solve this problem, we need to find the smallest positive integer $n$ such that the given conditions are satisfied for a regular $2016$-simplex grid with $n$ points on each edge. Let's break down the problem step by step.

1. Understanding the Problem:
- We have a $2016$-simplex grid with $n$ points on each edge.
- The points are labeled with real numbers.
- The sum of the labels on any line parallel to an edge of the simplex is zero.
- The labels are symmetric along any such line.
- Not all points are labeled with zero.

2. Lower Bound Analysis:
- Let's denote $t_{k,n}$ as the "tolerance" of a $k$-simplex with $n$ points per edge. This tolerance represents the side length of the largest $k$-simplexes you can place at each vertex of the original $k$-simplex such that all the points on those $k$-simplexes are zero, but not all points in $S$ are zero.
- For example, for a $2$-simplex (triangle), we have:
$$t_{2,7} = 2 \quad \text{and} \quad t_{2,8} = 2$$
This means that for $n = 7$ or $n = 8$, the tolerance is 2.

3. Inductive Hypothesis:
- Suppose the minimum $n$ for a $k$-dimensional simplex is $n_k = k^2 + k + 1$.
- We need to show that for a $(k+1)$-simplex, the minimum $n$ is $n_{k+1}$.

4. Inductive Step:
- For a $k$-simplex with $n_k$ points per edge, the tolerance $t_{k,n_k}$ is $n_{k-1} - 1$.
- For $n_{k+1}$, we need to ensure that:
$$t_{k,n_k+k+2} = t_{k,n_k+k+1} = n_{k-1} + k - 1$$
- For the $(k+1)$-simplex, consider the large $k$-simplex with $k+1$-th coordinate $n_{k+1} - n_k - k - 2$, which will have tolerance $n_{k-1} + k - 1$.
- Since the $k$-simplex with $k+1$-th coordinate 0 must have tolerance at least $n_k - 1$, we must have:
$$n_{k+1} - n_k + n_{k-1} - 3 \ge n_k - 1$$
Simplifying this inequality:
$$n_{k+1} \ge (2k^2 + 2k + 2) - (k^2 - k + 1) + 2 = k^2 + 3k + 3$$

5. Conclusion:
- By induction, the smallest $n$ for a $2016$-simplex is:
$$n_{2016} = 2016^2 + 2016 + 1 = 4066273$$

The final answer is $\boxed{4066273}$.