Given an integer $n > 5$, there are $n$ points in the plane with no three collinear. Tom sequentially erases a point each day until only three points remain. On the $i$-th day ($1 < i < n-3$), he notes a positive integer $v(i)$ representing the number of vertices in the current convex hull. Finally, $v(n-2) = 3$ when only three points remain. We aim to find the greatest possible value of the expression:

$$|v(1)-v(2)|+ |v(2)-v(3)| + \ldots + |v(n-3)-v(n-2)|.$$

### Solution Approach

1. Initial Setup:
- Initially, the convex hull can have at most $n$ vertices.
- Reducing the number of points step by step affects the vertices of the convex hull.

2. Understanding Convex Hull Changes:
- Removing a point from inside the convex hull does not change the number of vertices.
- Removing a point from the boundary reduces the vertex count by at least 1.

3. Maximizing the Expression:
- Begin with the maximal convex hull having all $n$ points as vertices, i.e., $v(1) = n$.
- Gradually remove the points strategically so that the convex hull loses its vertices one by one, ideally decreasing the vertex count by 1 each day.
- You will thus achieve a maximum change in the convex hull vertices each day, resulting in the expression $|v(i) - v(i+1)| = 1$ maximized wherever possible.

4. Expression Calculation:
- The sequence of vertex counts could be as simple as decreasing the hull by 1 vertex per day: $n, n-1, n-2, \ldots, 4, 3$.
- The expression becomes:
$$|(n) - (n-1)| + |(n-1) - (n-2)| + \ldots + |4 - 3|$$
- The number of terms in the expression is $n - 4$, with each term equaling 1, giving a sum:
$$(n-4) \times 1 = n-4$$

5. Ensuring Maximum Value:
- Each day except the very last when $3$ vertices are expected, has differences yielding $1$, ensuring maximum configuration is used.
- Subtract $1$ for each day's reduction starting at $n$ until reaching $v(n-2) = 3$.

Thus, the greatest possible value that the expression can obtain is:

$$\boxed{2n - 8}$$