PK 2017 Q21

Each ‘triple’ consisting of a cell and the two cells immediately below can have at most two odds (for if the bottom two are both odd, the one above is even, so they cannot be all odd). The whole diagram can be dissected into six of these (shaded) triples as shown in the top diagram, with three other (white) cells left over. These six triples have at most \(6 \times 2 = 12\) odds between them. Moreover, the three remaining white cells cannot all be odd; if we assign the values \(A\) and \(C\) to the lowest of these white cells, and \(B\) to the cell between them, then the cells above have values \(A + B\) and \(B + C\). The top white cell then contains \(A + 2B + C\), which is even when \(A\) and \(C\) are both odd. Hence the three white cells have at most two odds, giving the whole diagram at most 12 + 2 = 14 odds.
The second diagram shows one possible way of achieving this maximum of 14 odds.