League Lunacy

no player won a game by more than 5 points

This means that the minimum Points Won by any player was 6.

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There are 2 things that we can determine from the starting grid with a certainty:

• Andy's Score of 8 can only have been as a result of 2 Wins and 2 Draws:
  • 3 Wins would have been a Score of at least 9 (3+3+3+?>=9), and 1 Win would have been a Score of at most 6 points (3+1+1+1=6)
  • Thus:
    • Andy's Points Won = $(2 \times 11)+(2 \times 10)+(0 \times 0)=42$
• Chris had 0 Wins, 1 Draw and 3 Losses:
  • 0 Wins: If Chris Won just a single game, this would have left 17 Points Won (28-11) from his remaining 3 games - As each player got at least 6 points in every game, this is not possible.
  • 1 Draw:
    • Chris must have at least 1 Draw, because Andy has 2 Draws. With Bob and Eddie having 0 Draws, this leaves Chris and Dan, and he cannot Draw against the same person twice (as they play each other only once).
    • If Chris Drew 2 games, this would have left 8 Points Won from his remaining 2 games - As each player got at least 6 points in every game, this is not possible.
  • 3 Losses: No Wins and 1 Draw forces this to be 3.
  • Thus:
    • Chris's Points Lost = $(1 \times 10)+(3 \times 11) = 43$
    • Chris's Score = $(0 \times 3)+(1 \times 1)+(3 \times 0) = 1$

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Next, we can determine the following for Dan:

• Dan Drew 1 game:
  • 1 Draw for Chris means that Andy's 2nd Draw must have been with Dan - because Bob and Eddie already have 0 Draws.

Thus:

• Dan Won 3 games:
  • With Dan Drawing 1 game, this makes up 10 of his 32 Points Lost - leaving 22 Points Lost from his 3 remaining games.
  • No Losses: If Dan Lost a single game, this would have left 11 Points Lost from his remaining 2 games - As each player got at least 6 points in every game, this is not possible.
  • Thus:
    • Dan's Points Won = $(3 \times 11)+(1 \times 10)+(0 \times 0)=43$
    • Dan's Score = $(3 \times 3)+(1 \times 1)+(0 \times 0)=10$

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Now we can start narrowing down the Wins and Losses for Bob and Eddie:

• Both Bob and Eddie must have had a Loss at the hand of Andy, because Andy Drew against Chris and Dan.
• Everyone except Andy suffered a Loss against Dan.
• Either Bob or Eddie must have gained a Win in their game, and the other suffered a Loss.
• Thus (out of Bob and Eddie):
  • One of them has 1 Win and 3 Losses, and the other has 2 Wins and 2 Losses.

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With Bob's Points Lost being 37, we can now determine the following for Bob and Eddie:

• Bob got 2 Wins and 2 Losses:
  • As determined earlier; One of them has 1 Win and 3 Losses, and the other has 2 Wins and 2 Losses
  • 2 Losses: 3 Losses would have been 33 Points Lost, leaving 4 Points Lost from the remaining game - As each player got at least 6 points in every game, this is not possible.
  • Thus:
    • Bob's Score = $(2 \times 3)+(0 \times 1)+(2 \times 0)=6$
• Eddie got 1 Win and 3 Losses:
  • As a result of Bob and Eddie's juxtaposition, Bob got 2 Wins and 2 Losses, so Eddie must have got 1 Win and 3 Losses.
  • Thus:
    • Eddie's Score = $(1 \times 3)+(0 \times 1)+(3 \times 0)=3$

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The only things we are missing now are Bob's Points Won, and Eddie's Points Lost. Due to the fact that the sum of the Points Won column and the sum of the Points Lost column must be equivalent, if we find one, we can find the other.

Here, Eddie's Losses and Chris's Points Won are the key.

• Chris's Points Won in his Losses was 6 for each game (from 28 Points Won):
  • 1 Draw vs Andy leaves 18 points in his 3 Losses, meaning each Loss was exactly 6 Points Won for Chris.
• Eddie's Points Lost is 39:
  • 3 Losses is 33 Points Lost.
  • 1 Win vs Chris was 6 Points Lost.
  • Thus:
    • Bob's Points Won = Eddie's Points Lost = 39

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The sum of all the numbers in the completed table above is $\large\boxed{426}$