Tough one according to me. Your source? Anyway, I am giving a shot at it. I have no idea if I have covered all values but let us see.

Statement 1.
The sum of n and o is more than twice the sum of j and k.
Consider the set of values
1,6,7,8,9, 14 (there is a reason for selecting these values)
n + o = 14 + 9 = 23
1 + 6 = 7
So this set satisfies given condition.

In this case, median = 7 + 8 / 2 = 7.5
Mean = 45/6 = 7.5
So not larger.

In the above set, change 14 to 15 and mean > median. So with values satisfying given condition, we have 2 answers. Hence not sufficient.

Statement 2
The sum of k and o is 4/3 the sum of l and m
1,6,7,8,9, 14 works here and mean and median are equal.

Consider another set:
2,3 4, 5, 6 and 9.
Satisfies the condition, but here mean is 29/6 = 4.75 and median = 4.5
So for 2 sets of values, we get 2 answers. Hence this too is not sufficient.

Combining 1 and 2:
Here my values used in statement 2 help. Both values satisfy both conditions.
But we have already proven that we get 2 sets of answers.

So, if I have not gone wrong above, answer must be E. Feel free to point out if I have made any calc or judgment errors.

Wow - great explanation, vineeshp! You nailed it...really well done.

Just a couple things to add since I couldn't resist doing this problem once I saw it:

vineeshp selected great values here and did a really nice job explaining. I'm almost positive that he and I were thinking the same way, but just to clarify for strategy, I think what he probably did (and what I did) was to start with a simple set of numbers just to see how they react:

1,2,3,4,5,6,

here the median is the same as the mean (as with any set of consecutive integers) and statement 1 is already satisfied (5+6 > 2(1+2))

That gives us "no" to the original question. How do we get "yes"? Just bump up 6 to 7 and we add more weight to the high end of the mean without changing the median. Doing this it's pretty easy to get both answers and St. 1 is clearly insufficient.

Now with statement 2, we need to get a median that is divisible by 3 so that k + 0 can be 4/3 of it. If we use 4 and 5, that's 9, so k + 0 would be 12:

2, 3, 4, 5, 6, 9

Here you don't even need to do the math to recognize that we're adding value on the upper end of the range but not on the lower. Were the terms all consecutive mean would = median. But since we skip a few numbers on the right hand side, we're adding to the mean without changing the median, so the mean is greater.

BUT look at what didn't change - the lower limit (j, which we're calling 2 here). If we want to reduce the mean, we just have to reduce j, and we're not constrained there at all. So we can quite easily pull j down as far as we'd like to*. So we can definitely get "no" to the overall question here, too.

And taking both statements together you should see that we don't at all change our ability to reduce the lower limit. Statement 1 only requires that n + 0 be more than twice j + k, so we're allowed to reduce j all we want. So without even really calculating anything, we can just play with the upper and lower limits of the set to determine that the answer is E.

*Note that the question does require positive integers, so technically we can't reduce j to a negative number. But since it's easy to use 1-6 as our starting point that's what I did - notice however that we could just as easily have used 101-106 without changing the relationship, and at that point we'd have had a huge buffer to reduce the lowest term in the set, so the logic still holds.

Hey Brian!

Thanks for your kind words.

Yes, you are right, as soon as I saw 6 unequal positive integers, I took 1-6 and verified how it responds to statement 1. Once I realized the way to go, I moved the values a little high so that I had sufficient buffer to reduce the lowest value.

The only reason I did not mention that here is because of the effort in typing out the whole thing again while transferring from paper to PC.