6. Given a convex quadrilateral ABCD. Point $M$ is the midpoint of side BC, and point $N$ is the midpoint of side CD. Segments AM, AN, and $MN$ divide the quadrilateral into four triangles, the areas of which, written in some order, are consecutive natural numbers. What is the maximum possible area of triangle $\mathrm{ABD}$?
Answer: 6.

Solution. Estimation. Let $n, n+1, n+2, n+3$ be the areas of the four triangles. Then the area of quadrilateral $ABCD$ is $4n+6$. $MN$ is the midline of triangle $BCD$, so $S_{BCD} = 4S_{MCN}$, but $S_{MCN} \geq n$, hence $S_{BCD} \geq 4n$. Then $S_{ABD} = S_{ABCD} - S_{BCD} \leq 6$.

Example. If $ABCD$ is an isosceles trapezoid with bases $BC=4$ and $AD=6$ and height 2, then $S_{ABD} = 6$. The areas of triangles $CMN$, $ABM$, $AND$, and $AMN$ are 1, 2, 3, and 4, respectively, i.e., they are consecutive natural numbers.

Criteria. A fully correct solution - 7 points.

A correct solution, but the example is not shown to be suitable (the fact that the areas of the four specified triangles are consecutive natural numbers is not justified) - 5 points.

It is proven that the area does not exceed 6, but there is no example where the area is 6 (or it is incorrect) - 4 points.

An example is obtained where the area of triangle $ABD$ is 6 (with justification and verification of calculations), but there is no explanation of why it cannot be more than 6 - **3** points.

The example is correct, but without justification that it fits the conditions of the problem (that the areas of the required triangles are 4 consecutive numbers) - **1** point.