To answer your question, YES, if the length of an edge is the same in each direction, then starting at a vertex $y_1$ and picking the next edge of maximum length that visits a new vertex will return a Hamiltonian path that is at least as long as the Hamiltonian path returned by starting at a vertex $u_1$ picking the next edge of minimum length that visits a new vertex, for any choices of the starting vertices $y_1$ and $u_1$ in the complete graph. If the starting vertices are the same, then picking the next edge of maximum length that visits a new vertex will return a Hamiltonian path that is strictly longer than the Hamiltonian path returned by picking the next edge of minimum length that visits a new vertex.
To elaborate: Let $K_N$ be a complete graph with a length function $\ell$ on the edges of $K_N$ that satisfies the following:
No two edges incident to the same vertex are of the same length.
$\ell(vw)=\ell(wv)$ for every pair of vertices $w,v \in K_N$.
Let $u_1$ be a vertex in $K_N$ and let $P^{\min}(u_1)$ be the Hamiltonian path $u_1u_2\ldots u_N$ on the complete graph $K_N$ chosen as follows: For each $j$, the edge $u_ju_{j+1}$ is the edge with the smallest length incident to $u_j$ and not in $\{u_ju_1, \ldots, u_ju_{j-1}\}$. Let $y_1$ be a vertex in $K_N$ and let $P^{\max}(y_1)$ be the Hamiltonian path $y_1y_2 \ldots y_N$ on the complete graph chosen as follows: For each $k$, the edge $y_ky_{k+1}$ is the edge with the biggest length incident to $y_k$ and not in $\{y_ky_1, \ldots, y_ky_{k-1}\}$.
THM 1: Then there is a complete matching $M$ that matches each edge $u_ju_{j+1} \in P^{\min}(u_1)$ with an edge $M(u_ju_{j+1}) \doteq$ $y_ky_{k+1} \in P^{\max}(y_1)$ that satisfies the following: $\ell(u_ju_{j+1}) \le \ell(M(u_ju_{j+1})) = \ell(y_ky_{k+1})$. In particular, the inequality $\ell(P^{\min}(u_1)) \le \ell(P^{\max}(y_1))$ for any 2 vertices $u_1,y_1 \in K_N$, where, in a slight overload of notation, $\ell(W) \doteq \sum_{i=1}^{r} \ell(v_iv_{i+1})$ for any walk $W=v_1v_2 \ldots v_rv_{r+1}$ in $K_N$.
First, for each integer $j$ let $\pi(j)$ be the integer $k$ such that the $k$-th vertex $y_k$ of $P^{\max}(y_1)$ is the $j$-th vertex $u_j$ of $P^{\min}(u_1)$.
We construct $M$ for THM 1 as follows: We first find $M(u_{N-1}u_{N})$, and then $M(u_{N-2}u_{N-1})$, and so on. So let us assume that we have found $M(u_{N-j'}u_{N-j'+1})$ for each positive integer $j'<j$. We now find $M(u_{N-j}u_{N-j+1})$
Case 1: If there exists a nonnegative integer $j' < j$ such that $\pi(N-j') > \pi(N-j)$, put $M(u_{N-j}u_{N-j+1})$ to be the edge $y_{\pi(N-j)}y_{\pi(N-j)+1}$. We claim the following: $$M(u_{N-j}u_{N-j+1}) \doteq \ell(y_{\pi(N-j)}y_{\pi(N-j)+1}) \ge \ell(u_{N-j}u_{N-j+1}).$$ [Indeed, let us write $N-j'\doteq a$ where $j'$ is as above, and let us write $N-j \doteq b$. Then both $a>b$ and $\pi(a)>\pi(b)$. Then $\pi(b)<\pi(a)$ and the construction of $P^{\max}(y_1)$ gives $\ell(y_{\pi(b)}y_{\pi(b)+1}) \ge \ell(y_{\pi(b)}y_{\pi(a)})$ Also, $b<a$ and the construction of $P^{\min}(u_1)$ gives $\ell(u_bu_{b+1}) \le \ell(u_bu_a)$. But $y_{\pi(b)}=u_b$ and $y_{\pi(a)}=u_a$ so $\ell(u_au_b)$ $=$ $\ell(y_{\pi(a)}y_{\pi(b)})$. So putting this together gives $$\ell(y_{\pi(b)}y_{\pi(b)+1}) \ge \ell(y_{\pi(b)}y_{\pi(a)}) = \ell(u_au_b) \ge \ell(u_bu_{b+1})$$ Note however that $\ell(u_bu_{b+1})=\ell(u_{N-j}u_{N-j+1})$ [because $N-j=b$] while $\ell(y_{\pi(b)}y_{\pi(b)+1})= M(u_{N-j}u_{N-j+1})$ and so this gives indeed $\ell(y_{\pi(N-j)}y_{\pi(N-j+1)})$ $\ge \ell(u_{N-j}u_{N-j+1})$, which is as desired.]
Case 2: If there does not exist a $j'$ as in Case 1. Then the inequality $\pi(N-j) > \pi(N-j')$ for all integers $j' < j$, so let us now set $j''$ to be the integer in $\{0,1, \ldots , j-1\}$ so that $\pi(N-j'')$ is the largest out of $\pi(N),\pi(N-1), \ldots \pi(N-j-1)\}$. Then $\pi(N-j'')$ is the second largest out of $\pi(N),\pi(N-1), \ldots \pi(N-j)$. So set $M(u_{N-j}u_{N-j+1})$ to be the edge $y_{\pi(N-j'')}y_{\pi(N-j'')+1}$. We claim the following: $$M(u_{N-j}u_{N-j+1}) \doteq \ell(y_{\pi(N-j'')}y_{\pi(N-j'')+1}) \ge \ell(u_{N-j}u_{N-j+1}).$$ [Indeed, let us write $N-j'' \doteq a$ and $N-j \doteq b$. Then $a>b$ while $\pi(a) < \pi(b)$. Then $\pi(a) < \pi(b)$ and the construction of $P^{\max}(y_1)$ implies $\ell(y_{\pi(a)}y_{\pi(a)+1}) \ge \ell(y_{\pi(a)}y_{\pi(b)})$ in $K_N$. Also $b<a$ and the construction of $P^{\min}(u_1)$ implies $\ell(u_bu_{b+1})$ $\le$ $\ell(u_bu_a)$. However, $u_a=y_{\pi(a)}$ and $u_b=y_{\pi(b)}$ so $\ell(u_au_b)$ $=$ $\ell(y_{\pi(a)}y_{\pi(b)})$. So putting this together gives $$\ell(y_{\pi(a)}y_{\pi(a)+1}) \ge \ell(y_{\pi(a)}y_{\pi(b)}) = \ell(u_au_b) \ge \ell(u_bu_{b+1})$$ Note however that $\ell(u_bu_{b+1})=\ell(u_{N-j}u_{N-j+1})$ [because $N-j=b$] while $\ell(y_{\pi(a)}y_{\pi(a)+1})= M(u_{N-j}u_{N-j+1})$ [because $N-j''=a$] and so this gives indeed $\ell(y_{\pi(N-j'')}y_{\pi(N-j'')+1})$ $\ge \ell(u_{N-j}u_{N-j+1})$, which is as desired.]
We can check that $M$ as defined above is indeed a matching as well, and so THM 1 follows.
From this we conclude that the inequality $\ell(P^{\min}(u_1)) \le \ell(P^{\max}(y_1))$ holds for all $u_1,y_1 \in V(K_N)$. $\surd$
We note that THM 1 holds if even if Condition 1. above for the length function $\ell$ does not hold.
We note that we cannot necessarily make the inequality in THM 1 strict. To elborate, there are instances where $\ell(P^{\min}(u_1))=\ell(P^{\max}(y_1))$ for some $u_1,y_1 \in K_N$ even with Condition 1. above for the length function $\ell$ holding. Indeed, take $N=3$ and the graph $K_3$ on $\{y,w,v\}$ where $\ell(yw)=100$; $\ell(yv)=99$ and $\ell(wv)=1$. Then $\ell(P^{\max}(y))= \ell(ywv) = \ell(P^{\min}(v))=\ell(vwy)$.
However if $u_1=y_1=v$ then [and condition 1. for $\ell$ holds] then the strict inequality $\ell(P^{\min}(v))< \ell(P^{\max}(v))$ holds for all $N \ge 3$; to see this note that $M$ as constructed above will map $u_1u_2=vu_2$ to $y_1y_2=vy_2$ and $\ell(u_1u_2) < \ell(y_1y_2)$.
We note that the condition that $\ell(wv)=\ell(vw)$ is a necessary one for THM 1 to hold; otherwise take the graph $K_3$ on $\{y,w,v\}$ where $\ell(yw)=100$; $\ell(wy)=1$; $\ell(yv)=\ell(vy)=1$ and $\ell(wv)=\ell(vw)=2$. Then $P^{\min}(v) = vyw$ and has length $1+100=101$ whereas $P^{\max}(v)=vwy$ has length $2+1=3$.
