Amongst any 5 natural numbers there are 3 such that the sum of them is divisible by 3. Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that the sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisible by 3. It is easy to see that amongst any 9 natural numbers there are 5 such that the sum of them is divisible by 5. Since we have 9 numbers, all of them are divisible by 3, there are 5 such that the sum of them is divisible by 15.

## $\operatorname{COM} 4$.

$n$ points are given in a plane, no three of them collinear. One observes that no matter how we label the points from 1 to $n$, the broken line joining the points $1,2,3, \ldots, n$ (in this order) does not intersect itself.

Find the maximal value of $n$.

Solution. Notice that $n=4$ satisfies the condition. Indeed, for a

concave quadrilateral, this can be checked immediately.

Then, observe that for $n \geq 5$ one can choose four points $A, B, C, D$ such that $A B C D$ is a convex quadrilateral. The diagonals $A C$ and $B D$ intersect at a point, hence labeling $A, B, C, D$ with $1,2,3,4$ we reach a contradiction.

Thus, it is sufficient to prove that from five points we can select four that are vertices of a convex quadrilateral. Consider the convex hull of the five points set. If this is not a triangle we are done. If it is a triangle, then draw the line through the two points inside the triangle. This line meets exactly two sides of the triangle. Let $A$ be the common vertex of these sides. Then the four remaining points solve the claim.