It's trivial to find a "near-optimum" result in reasonable time
I've never played dart outside of the casual pub going, so I don't know any of the real intricacies of the game. However, it seemed intuitive to me that, when designing the layout of the dart board, we'd want as high of a penalty for missing a shot as possible -- that is, we'd want to maximize the difference between neighboring shots.
But I'm a 19th century dart player. I don't care much for mathematical proofs, or rigid brute force optimizations. Without thinking too deeply about it, I just... gave it a go?
First try
I started out by trying to just interleave the available numbers in reverse order: every second number would decrement from twenty, the other numbers would increment from 1. This gave the sequence of 20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 10
.
Both intuitively and by applying the measurement from the question's linked article, it was obvious that this wasn't very well balanced. Intuitively is seemed like the penalty would decrease as we moved towards the end of the sequence. The given measurement describes this in a slightly less intuitive way: the trios that started with a high square (i.e. those with the "small" numbers in the middle) would decrease as we moved towards the end of the sequence. The trios that started with a low square (i.e. those with the "big" numbers in the middle) would increase as we moved towards the end. This indicates a smaller variation overall near the end of the sequence than at the start.
Regardless, this sequence didn't feel nice, which just wouldn't do. No rigid mathematical proof or anything. I want to throw pointy sticks, not think more than I need to.
Second try
For the second go I tried to mitigate that by placing numbers that are close to eachother as far apart as I could. At every second step I tried to find a number that "balanced" the previous trio of numbers as well as possible (i.e. 20-1-? would have 10 added as the last number), but wasn't particularly rigid about it.
The process looked a little like this:
Index: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Sequence: 20 __ __ __ __ __ __ __ __ __ 19 __ __ __ __ __ __ __ __ __
Sequence: 20 1 __ __ __ __ __ __ __ __ 19 2 __ __ __ __ __ __ __ __
Sequence: 20 1 10 __ __ __ __ __ __ __ 19 2 9 __ __ __ __ __ __ __
Sequence: 20 1 10 18 __ __ __ __ __ __ 19 2 9 17 __ __ __ __ __ __
Sequence: 20 1 10 18 3 __ __ __ __ __ 19 2 9 17 4 __ __ __ __ __
Sequence: 20 1 10 18 3 8 __ __ __ __ 19 2 9 17 4 11 __ __ __ __
# the thinking starts becoming less rigid, since no real good numbers are available any longer
# I try to just... balance things out as I feel like
Sequence: 20 1 10 18 3 8 16 __ __ __ 19 2 9 17 4 11 12 __ __ __
Sequence: 20 1 10 18 3 8 16 __ __ __ 19 2 9 17 4 11 12 5 __ __
Sequence: 20 1 10 18 3 8 16 __ __ __ 19 2 9 17 4 11 12 5 13 __
Sequence: 20 1 10 18 3 8 16 __ __ __ 19 2 9 17 4 11 12 5 13 6
Sequence: 20 1 10 18 3 8 16 7 __ __ 19 2 9 17 4 11 12 5 13 6
Sequence: 20 1 10 18 3 8 16 7 15 __ 19 2 9 17 4 11 12 5 13 6
Sequence: 20 1 10 18 3 8 16 7 15 14 19 2 9 17 4 11 12 5 13 6
This gives a final sequence of 20 1 10 18 3 8 16 7 15 14 19 2 9 17 4 11 12 5 13 6
. Notably I didn't even try to optimize it after placing all the numbers -- this layout felt good enough, so I wouldn't want to spend any more time not throwing arrows at walls.
The scoring mechanism from the article linked in the question gives this a sum of squares of
$$(20 + 1 + 10)^2 + (1 + 10 + 18)^2 + \dots + (13 + 6 + 20)^2 + (6 + 20 + 1)^2 = 20410$$
Which makes it even more optimal than the "London" layout, according to that measure.
So... by just doing it, I think it's become clear that reaching a layout that's at least as "near-optimum" as the classic one is possible by hand ;)
Reframe
Of course, this is only "near-optimum" if we consider a) the measurement used in the article to be a valid approximation of what we want from a dart board, and b) we consider the $20478$ score of the London arrangement to be "near-optimum".
Given that the article goes on to find many, many arrangements that have scores closer to $19950$, I'd challenge that last point. As such, I don't think there's very much value in the question, given that the London arrangement (and mine) isn't near-optimum.