The existing scenario is as follows.
[The key move herein is that, for instance, $1+2+3+4+5+6$ can be solved by adding the numbers in pairs -- 1+6, 2+5, 3+4 -- giving $3*7$. If the count is odd, add the middle number as a special case.]
1 to each = $1+2+3...+8$ = $4*9$=36.
9 is the same.
2 to each is 1 + $1+2+...+7$ = $(3*8)+4 + 1$ = 29.
8 is the same.
3 to each is 2+1 + $1+2+...+6$ = $(3*7) + 1+2$ = 24.
7 is the same.
4 to each is 3+2+1 + $1+2+...+5$ = $(2*6)+3 + 1+2+3$ = 21.
6 is the same.
5 (being the middle) is $4+3+2+1+1+2+3+4$ = $2*(2*5)$ = 20.
+= $2*36+ 2*29+ 2*24+ 2*21 +20$ = $2*(36+29+24+21) +20$ = 240.
[If it was very many stations, one would generate a formula... again treating the odds and evens differently.]
It is fairly obvious that adding the new connection near the existing middle one -- from 4 to 6 -- will have less effect than adding it further away from the existing middle one.
At the other end... the, or a, limit case is connecting 1 to 9 -- completing the/a loop. That gives 9 instances of (1+2+3+4+4+3+2+1) , which = $9*( 4*(5) )$ = 180. [Any journey more than 4 should be done in the opposite direction.]
The question then is that of whether the foregoing is the best case, or not. (If it is not, the best case will (presumably) be around the middle (of the halves) somewhere.)
It it is the best case, the next connection would be worse. [If, conversely, the next connection is better, then we have to work out where the sweet spot is.]
Connecting 2 to 9 gives us a loop, again -- 8 instances of $1+2+3+4+3+2+1$ = 104 -- plus (for station 1) $1+2+2+3+3+4+4+5$ = 24. That is a total of 128... which is better than the simple loop ... so it is improving... so we have to work out where the sweet spot is.
The question was whether or not we have to solve this puzzle just by brute force.
I daresay that the remainder of the working-out would be apparent to someone who had been following so far. (No doubt there are some maths geniuses out there who would have already been able to see that the limit loop case was, or was not, optimal.)
If there was a large number of stations, we could work out the pattern as I have detailed, and find the limit case (somewhere around the middle) by eliminating corresponding elements [e.g. "$2*(1+1+2+2+3)$" appearing in two different candidate solutions] and just looking at the remaining items.
It happens that, to see the pattern, we have to work out (I suspect) at least 4 more variants (2 even and 2 odd). Thus, for this small example, we end up doing most of it by brute force, in working out the pattern.
For a large case, we would still have to work out the pattern, but we would solve the problem by looking at the pattern, without doing the calculations. [There would by many instances of "n" and "n/2" and "(n+1)/2", before simplifying. ...Also "nCr".]
On the one hand, I imagine that a professional statistician could work out the pattern without actually going through all of the above "manually" as I did. On the other hand, even a professional statistician has to get their head around a novel problem, to see what equations emerge.