Suppose the opposite. Colour the edges of the grid in the colours $1,2,3$ such that two edges sharing a vertex have the same colour iff the angle between them is $120\deg$. It is easy to see that this colouring exists ($3$ classes of the hexagons with same coloured edges) and that "walking" around the polygon creates either $1 \rightarrow 2 \rightarrow 3 \rightarrow 1 \rightarrow 2...$ or $1 \rightarrow 3 \rightarrow 2 \rightarrow 1 \rightarrow 3...$ sequence (this depends on the direction we choose during the walk). It means that perimeter of the polygon is divisible by $3,$ contradicting the assumption.

IMO2021 wrote: Suppose the opposite. Colour the edges of the grid in the colours $1,2,3$ such that two edges sharing a vertex have the same colour iff the angle between them is $120\deg$. It is easy to see that this colouring exists ($3$ classes of the hexagons with same coloured edges) and that "walking" around the polygon creates either $1 \rightarrow 2 \rightarrow 3 \rightarrow 1 \rightarrow 2...$ or $1 \rightarrow 3 \rightarrow 2 \rightarrow 1 \rightarrow 3...$ sequence (this depends on the direction we choose during the walk). It means that perimeter of the polygon is divisible by $3,$ contradicting the assumption. Example of colouring:

If the perimeter isn't devided by 3,then whether P has a Internal or an external 120-degree angle or not?