The trap in this problem is built into the word median. The natural reflex for "make one location as profitable as possible" is "make every other location as small as possible and pour the rest into that one." That reflex is correct when you are maximizing the single largest value, but here you are maximizing the median, and the median is a rank, not a free magnitude. It is the fourth of the seven profits when they are lined up in order, so three locations must rank above it. You cannot make those three small, because each of them has to be larger than the median itself.
Line up the seven distinct profits from smallest to largest. The median is the fourth one, call it M. To push M as high as possible, work both sides toward their limits. The three locations below the median should be as small as distinct positive whole numbers allow, which is 1, 2, and 3, costing only 6 of the 56 million. The three locations above the median cannot be smaller than the median, and they must be distinct, so the cheapest they can be is M + 1, M + 2, and M + 3. Everything is now pinned to M, so add it all up: 1 + 2 + 3 + M + (M + 1) + (M + 2) + (M + 3) = 4M + 12, and that has to equal 56. Solving gives 4M = 44, so M = 11.
Check that 11 is actually reachable and that nothing higher is: the set 1, 2, 3, 11, 12, 13, 14 uses seven different whole numbers, sums to 56, and has 11 as its middle value. Try to push the median to 12 and even the cheapest legal set, 1, 2, 3, 12, 13, 14, 15, already sums to 60, which overshoots 56. So 11 is the greatest possible median, which is choice (C).
The dominant wrong answer is (E) 35, and it is exactly what the "maximize one, minimize the rest" reflex produces. Shrink six locations to 1 through 6, which is 21 million, and dump the remaining 35 million into the seventh. That does maximize a location, but the 35 million location is the single highest earner, not the median; the median is still the fourth-ranked location, which in that set is only 4. The reflex optimized the wrong variable. Choice (A) 4 is what you get if you slide the same effort to the opposite end and accidentally minimize the middle location instead of maximizing it, answering the least possible median when the greatest was asked. Choice (B) 8 is 56 ÷ 7, the value you would get if the seven profits were equal, which treats the median as if it were the average and ignores both the ranking and the requirement that the values be distinct. Choice (D) 14 is 56 ÷ 4, from a solver who correctly sees that some locations must rank above the median but then divides the whole 56 million among only the top four locations and forgets that the bottom three still have to take their floor of 1, 2, and 3.
The reusable method: to maximize a value that is defined by its rank, identify which other values its rank forces to be larger or smaller, push those to their tightest legal floor, push the freely shrinkable values to their floor too, set the total to the given sum, and solve. The key discipline is to ask which constraint actually binds at the extreme, rather than reaching for the generic "minimize everything else" move, because that move maximizes the extreme element instead of the ranked one.