Minimum number of times monkey must travel
There was an answer here a few hours ago claiming $n^2/4.$ I did not read it in full, and I do not know why it's been deleted. But I think the claim is correct. The minimum number of travels is ${n^2\over 4}$ for even $n$, and ${n^2-1 \over 4}$ for odd $n$.
(Credits: the necessity argument I stole from the now-deleted post. The sufficieny part is my own - I didn't get a chance to read that part of the deleted post and have no idea if its argument is same as mine.)
Necessity: For each $a_i$ among the $\lfloor n/2 \rfloor$ nodes in the first half, and each $a_j$ among the $\lceil n/2 \rceil$ nodes in the second half, there is one jump $(a_i, a_j)$. None of these jumps can share the same travel. So the number of travels $\ge \lfloor n/2 \rfloor \lceil n/2 \rceil = {n^2 \over 4}$ (if $n$ is even) or ${n^2 - 1 \over 4}$ (if $n$ is odd).
Sufficiency, via explicit construction:
In round $1$, we do all jumps of the form $(a_1, a_i)$ and $(a_i, a_n)$. These can be assembled into $n-1$ travels as follows: the direct jump $(a_1, a_n)$, and, the two-hop path $((a_1, a_i), (a_i, a_n))$ for all $i \in [2,n-1]$. Since there are $n-2$ choices for $i\in [2,n-1]$, plus the direct jump, the total number of travels is $n-1$.
In round $2$, we do all jumps of the form $(a_2, a_i)$ and $(a_i, a_{n-1})$. By the same argument as above, there are $n-4$ choices for $i \in [3, n-2]$, plus the direct jump $(a_2, a_{n-1})$, for a total of $n-3$ travels. (Obviously, for each travel we need to attach $(a_1, a_2)$ at the front and $(a_{n-1}, a_n)$ at the back.)
In general, in round $k$, we do all jumps of the form $(a_k, a_i)$ and $(a_i, a_{n-k+1})$, for a total of $n-2k+1$ travels. (Obviously, for each travel we need to attach $(a_1, a_k)$ at the front and $(a_{n-k+1}, a_n)$ at the back.)
We keep doing this as long as $k < n-k+1$.
For even $n$, this means $k \le {n \over 2}$ (entire first half), and $n-k+1 \ge {n \over 2} + 1$ (entire second half).
For odd $n$, this means the midpoint ${n+1\over 2}$ is never either $k$ or $n-k+1$. We have: $k \le {n+1 \over 2} - 1$ (entire first half, exclude midpoint), and $n-k+1 \ge {n+1 \over 2} +1$ (entire second half, exclude midpoint).
Claim: After we do this for all rounds, all jumps have now been used.
Proof: consider any jump $(a_p, a_q)$. Either $p$ is in the first half (exclude midpoint if $n$ odd), or $q$ is in the second half (exclude midpoint if $n$ odd). Thus, this jump has been used in some round. Note that if $n$ is odd and $p$ or $q$ is the midpoint, the argument still applies to the other point.
Summing up all rounds, the total number of travels $=f(n) = (n-1) + (n-3) + (n-5) + \dots$
Even $n=2m: f(n) = (2m-1) + (2(m-1)-1) + \dots + 5 + 3 + 1 = m^2 = {n^2 \over 4}$.
Odd $n=2m+1: f(n) = 2m + \dots + 4 + 2 = 2(m + \dots + 2 + 1) = m(m+1) = {n^2 -1 \over 4}$.
There is a relatively straightforward lower bound.
Look at the total length of all the jumps that the monkey needs to do. There is just $1$ jump of length $n-1$ (going from $a_1$ to $a_n$), there are $2$ of length $n-2$, $3$ of length $n-3$, etc. The total length $L$ of all the jumps is therefore: $$L = \sum_{i=1}^{n-1} {i(n-i)} = \frac{(n-1)n(n+1)}{6}$$
Each traversal of the path is of length $n-1$, so the number of traversals T is at least $L/(n-1)$: $$T = \frac{L}{n-1} = \frac{n(n+1)}{6}$$
This is only a simple lower bound, and can only be reached for very small values of $n$.
The problem is that there are only $n-1$ possible jumps that start from $a_1$, and every traversal must include one. For larger $n$ you obviously have $T>n-1$, so the monkey will have to repeat jumps. Also, there is only one jump landing on $a_2$, but there are $n-2$ leaving it, so for $n>3$ there must again be repeat jumps.
This means that the lower bound gets more inaccurate for larger $n$, but I have not tried to quantify by how much.