The answer is $2n-8$

First, define $L=$$|v(1)-v(2)|+ |v(2)-v(3)| + \ldots + |v(n-3)-v(n-2)|$ Let us define $s_i$ to be the the value $n+1-i - v_i$. We will show that this value is non-increasing. To do this we first prove a claim. $Claim 1$: If $v_{i+1}+1\le v_{i}$, then $v_{i+1}=v_i -1$ Proof: If we pick one of the points in the interior of the hull, it is obvious that the $v_i$ remains unchanged, Now we consider what happens when one of the vertices is removed. It is sufficient to prove that the other vertices in the previous hull still remain in the new hull. This is true since there doesnt exist a triple of points which contain one of the previous vertices of the hull in its interior.(Contradicts previous hull). Now we show that $s_i$ is non-increasing. If $v_{i+1}=v_i -1$ then $s_{i+1}=s_{i}$, else if $v_{i}+l= v_{i+1}$, the $s_{i+1}=s_i - l -1$ Now we will find the maximal value of $L$.If $v_{i+1}=v_i -1$ we stay that this step is $decreasing$, else we call it $increasing$. If $v_{i+1}=v_i -1$ for all i, we get $L=n-3$. So from here on out we assume that the maximum number of decreasing steps is $n-4$. Let the increasing steps be $k_1,k_2,\hdots, k_l$ with $v_{k_i}+a_i=v_{k_i+1}$. So $L$ will be the sum of $a_i$ + the number of decreasing steps. Consider how $s_1$ changes, It wont change when we have a decreasing step but for an increasing step we get $s_{k_i+1}= s_{k_i}-a_i-1$ So in the end we will have $s_{n-2}=s_1 - a_1-a_2+\hdots - a_l - l$, Since $s_{n-2}=0$, we get $1 + a_1+a_2+\hdots +a_l \le a_1+a_2+\hdots + a_l+l \le s_1\le n-3 $. Therefore $L$ is at most $n-4+n-4=2n-8$ (sum of $a_i$+ at most $n-4$) The construction is rather straightforward. For a certain $n$ build an isosceles triangle with a convex $n-1$-gon using the triangles base and is completely inside it. Now on day one Tom erases the vertex opposite the triangles base and then he can erase in any order.