This is a logic problem, not a math problem. It's impossible to solve it with equations like others have tried. Also, the wording of the problem was very important.
2450 has 5 prime factors (2,5,5,7,7). This means that the ages must be broken up into products of these 5 numbers. Because there are 3 guests, the logical breakdown is that one of these guests is a child aged 2, 5, or 7, and the other two guests are the products of the remainders.
These are the possible breakdowns in this situation:
2 - [35,35] or [25,49] or [5,245] or [7,175]
5 - [10,49] or [14,35] or [2,245] or [5,98] or [7,70]
7 - [10,35] or [14,25] or [2,175] or [5,70] or [7,50]
Now for the second bit of info: the sum is 2 times Sara's age. Without knowing that age, the only thing this gives us is that the sum must be even. BUT the kicker here is that Sara says that this isn't enough information, so there must be 2 or more combinations that add to twice her age. Let's find the sums (in order, getting rid of unrealistic ages):
72,76
64,54,108,82
52,46,82,64
Duplicates are: 64, 82
So Sara is looking at one of these two combinations:
[5,10,49] or [7,7,50]
[5,7,70] or [7,5,70]
The second one are identical, so the duplicate is:
[5,10,49] or [7,7,50]
(Also, this means that Sara is 32)
So the last bit of info must let Sara know which is the valid combination. Eric says that he is older than all guests, and Sara knows the answer after that. This means that one of these combinations has a value older (or equal to) than Eric. This would be [7,7,50] as 50 is the oldest age.
This means that the guest combination is
[5,10,49]
So Eric is older than 49. BUT he must not be older than 50, or the previous cancellation wouldn't be valid.
THEREFORE:
Eric is 50.
QED