Premier League Predictions

The final score is $\sum_{n=1}^{20} |p_n-n| = \sum_{n=1}^{20} (\pm p_n \mp n),$ where $p_n$ is the prediction for the team that eventually finishes in $n$ th place. This is a signed sum with two copies of every integer from $1$ to $20,$ with $20$ pluses and $20$ minuses.

The maximum possible value of this signed sum comes when we put the pluses on the largest numbers: $20+20+19+19+\cdots+11+11 -10-10-9-9-\cdots-1-1 = 200.$

The other important point is that this signed sum is always even, because $|p_n - n| \equiv p_n-n \pmod 2,$ so $\sum_{n=1}^{20} |p_n-n| \equiv \sum_{n=1}^{20} (p_n-n) = \sum_{n=1}^{20} p_n - \sum_{n=1}^{20} n = 0,$ since the $p_n$ are a permutation of $1, \ldots, 20.$

So the final score must be an even number $\le 200,$ and $\fbox{114}$ is the only possible choice. To be complete, I should also show that $114$ is possible. Here's a sequence $p_n$ that should do the trick: $20,19,18,5,6,7,8,9,10,4,11,12,13,14,15,16,17,3,2,1.$ This gives a final score of $19+17+15+1+1+1+1+1+1+6+0+0+0+0+0+0+0+15+17+19 = 114.$

(Remark: I believe that every even number between $0$ and $200$ inclusive is a possible final score, but I don't have a short proof.)