Hamiltonian path greedy and anti-greedy algorithms
No.

In fact, we can prove the following stronger proposition.
Claim. Given a complete weighted undirected graph, a run of the greedy algorithm on it and a run of the anti-greedy algorithm on it, the $i$-th heaviest edge chosen by the greedy algorithm weighs no more than the $i$-th heaviest edge chose by the anti-greedy algorithm, for all valid $i$.
Proof. It is enough (and necessary) to show, given any weight $w$, the number of edges chosen by the greedy algorithm that weigh more than $w$ is no more than the number of edges chosen by the anti-greedy algorithm that weigh more than $w$.
Fix an arbitrary weight $w$. Let $\overrightarrow{u_1v_1}$, $\overrightarrow{u_2v_2}$, $\cdots$, $\overrightarrow{u_kv_k}$ be the edges chosen by the greedy algorithm that weigh more than $w$, in the same order as they were chosen by the greedy algorithm. Assume $k\ge1$; otherwise we are done.
Note that $u_1, u_2, \cdots, u_k$ and $v_k$ are distinct, since they are the starting vertex of distinct edges and the ending point of the last edge, respectively. Let $u_{k+1}=v_k$.
Consider any two vertex $u_p$ and $u_q$ such that the greedy algorithm visits $u_p$ earlier than $u_q$, i.e., $p&lt;q$. We know $\overrightarrow{u_pv_p}$ weighs no more than $\overrightarrow{u_pu_q}$ since it weighs the least among all edges from $u_p$ to unvisited nodes. Since $\overrightarrow{u_pv_p}$ weighs more than $w$, so does $\overrightarrow{u_pu_q}$. We have just proved the simple and critical observation that all edges among $u_1, u_2,\cdots,u_{k+1}$ weigh more than $w$.
The anti-greedy algorithm will visit each vertex in $u_1, u_2, \cdots, u_{k+1}$  soon or later, since it will find a hamilton path. Consider the $m$-th time when the anti-greedy algorithm have just visited such a node. If $m\lt k+1$, then there is at least another such node that has not been visited, which means the next edge it will choose must weigh no less than the edge to that unvisited node, which weighs more than $w$ as we just proved above. That is, for each of the first $k$ times the anti-greedy algorithm must choose an edge that weigh more than $w$.  $\quad\checkmark$