Divide the melons up into two groups: those weighing more than 6 kilograms, and those weighing 6 or fewer kilograms. Call them group $A$ and group $B$. How many melons can be in group $A$? Since each melon weighs more than 6 kilograms, there must be less than $\frac{270}{6}=45$ melons; that is, the number of melons in $A$ is at most 44. We can give 4 to each person and not overload them, because the weight of 4 melons is at most $4\cdot7=28$ kilograms. So suppose we have divided up the melons in group $A$ among the carriers. Now take some melon in group $B$. It has weight $x\leq6$ kilograms. Now suppose everyone is carrying more than $30-x$ kilograms already, so that nobody can carry $x$. That'd mean that the total weight is more than $11(30-x)+x=330-10x\geq270$. This is false, so someone is carrying at most $30-x$ kilograms, and we can give the melon weighing $x$ to this carrier. We are hence done - we can give all the melons to somebody without having anyone carry more than $30$ kilograms.