For third edition rules
Here's one way how:
The numbers show when tiles are placed. Black is the first player to move and always places a meeple where the number is. Blue moves second and never places a meeple.
Proof of optimality
Meeple usage optimality:
The first player to move will have 36 turns and thus only 36 opportunities to place meeples for points. The best usage of those 36 is:
- 7 for fields
- 6 for cloisters
- 16 for cities
- 7 for roads
No more can be placed on fields since you don't get those back. The number on cities and cloisters are sufficient to count all possible city/cloister points (except the odd city tile, see city optimality). So the only possible useful change would be to get more roads, which adds at most 2 because all roads without endpoints are already scored. But 2 is less than the amount we would lose by having less cities (each extra city adds 3 field points to each big field), cloisters (9 each), or small field (6 each).
There are 44 city tiles and 10 shields. 5 of the city tiles have two separate cities on them and so if you put them on separate cities they can be counted twice. This would all be worth 59 points and completed cities are worth double making 118. However all completed cities must have an even vertical perimeter and an even horizontal perimeter because they must be complete on both left/right and up/down (a diagonal city tile has 1 horizontal and 1 vertical perimeter, not 1 total perimeter). For example say a city is 5 tiles tall, then its vertical perimeter must be 10 if completed, yes you could have concavity but the same rules apply to that. And since the total horizontal+vertical perimeter of all available tiles is odd, you cannot complete all city tiles, the best you can do is leave out one tile. This means the maximum city score is
We need to maximize the number of completed cities since we can construct big fields that touch all cities. A completed city must have at least 4 tiles with a corner that do not have their own city touching them. Diagonal city tiles (i.e black #29) have 1 such corner and the simple side piece cities (i.e. the starting tile) have 2 such corners (all other city tiles have 0). There are 10 diagonals city tiles and 28 simple side piece cities (tiles with 2 simple side piece cities count as 2). This gives
10+28*2=66 corners which allows for a maximum of 16 cities.
Now to maximize the number of fields that touch these 16 cities. I will attempt to explain this intuitively but user1873 came to the same conclusion and proves it using graph theory (see their answer - which is the same in other ways except only has 15 cities). At most we can have 2 big fields touching all cities. A third field must be stuck inside these big fields and therefore only touching only 2 cities. In order to touch a third city it would have to cross one of the big fields, but this would split that big field into two smaller fields.
So we have 2 big fields touching 16 cities and 5 small fields touching 2 cities.
Each completed road has 2 end points. There are 32 road tiles without end points which can fit into any completed road. So the maximum points you can get from roads are
2 * #roads + 32. Since we only have time to place meeples on 7 roads this gives
Each of 6 cloisters can score a maximum of 9 for
Sum of optimal features
116+126+46+54 = 342
I have to say that this is a beautiful question because not only is the proof multi faceted with outside the box thinking required, but because actually finding a configuration that achieved it was really nontrivial and fun. So much so that even 6 years later I remember the joy of figuring out to put that gap in the city to allow it to complete and the tile missing to prevent the two big fields from joining. That beauty is why I've revisited this and rewritten this answer to make it more clear and also to break it down into an example with all moves and a separate proof.
An open question would be what the maximum possible score in a one player game is. You'd have 31 more turns to place meeples but the only possible points to grab would be completing more roads. There are 12 road endpoints unused in my example but how in the world would you rearrange them to connect with each other without breaking other things? Another open question is what the total collaborative score is for games of 2-5 players. This would be tricky since sharing cities, fields and even roads would need to be done as much as possible but it surely won't be possible to share everything. These questions will be an order of magnitude harder to solve and prove optimal.