Punishment number 2018!

Suppose we have chosen values $a_1, \dots, a_{18}$ , and the expression to be maximized has value $M$ . Consider what happens if we make the change $a_i \mapsto a_i-1$ and $a_j \mapsto a_j+1$ . The new value of $M$ will be $M' = M - a_{i-2}a_{i-1} - a_{i-1}a_{i+1} - a_{i+1}a_{i+2} + a_{j-2}a_{j-1} + a_{j-1}a_{j+1} + a_{j+1}a_{j+2}.$ This is an improvement ( $M' > M$ ) if at least one of the products $a_{j-2}a_{j-1}, a_{j-1}a_{j+1}, a_{j+1}a_{j+2}$ is positive and greater than the corresponding values near $a_i$ . In other words, we wish to have at least three non-zero values next to each other, and once we have this, it pays to increase them at the expense of other values in the list.

All of this suggests to make three neighboring values equal to $27/3 = 9$ , and everything else zero. These neighboring values should not be too close to the beginning or the end of the list; otherwise, some of the terms will be missing. A maximal solution is $(a) = (0, 0, 9, 9, 9, 0, 0, \dots, 0),$ with $M = 9^3 = \boxed{729}.$