Let $E$ and $F$ be the respective midpoints of $BC,CD$ of a convex quadrilateral $ABCD$. Segments $AE,AF,EF$ cut the quadrilateral into four triangles whose areas are four consecutive integers. Find the maximum possible area of $\Delta BAD$.
1. Let the areas of the four triangles formed by segments \( AE, AF, \) and \( EF \) be \( x, x+1, x+2, x+3 \). Therefore, the total area of the quadrilateral \( ABCD \) is:
   \[
   \text{Area of } ABCD = x + (x+1) + (x+2) + (x+3) = 4x + 6
   \]

2. Since \( E \) and \( F \) are the midpoints of \( BC \) and \( CD \) respectively, the area of \( \triangle BCD \) is divided into four smaller triangles by \( E \) and \( F \). The area of \( \triangle BCD \) is four times the area of \( \triangle ECF \) because \( E \) and \( F \) are midpoints.

3. Let the area of \( \triangle ECF \) be \( y \). Then the area of \( \triangle BCD \) is:
   \[
   \text{Area of } \triangle BCD = 4y
   \]

4. Since the areas of the four triangles are consecutive integers, we can assume that the smallest area is \( x \). Therefore, the area of \( \triangle BCD \) must be at least \( 4x \) because it includes the areas of \( \triangle ECF \) and the other three triangles.

5. The total area of the quadrilateral \( ABCD \) is \( 4x + 6 \). Since the area of \( \triangle BCD \) is at least \( 4x \), the remaining area, which is the area of \( \triangle BAD \), is:
   \[
   \text{Area of } \triangle BAD = (4x + 6) - 4x = 6
   \]

6. Therefore, the maximum possible area of \( \triangle BAD \) is \( 6 \).

The final answer is \( \boxed{6} \).