Matrix Math - Maximizing
As you pointed out already, the combinatorics are not on your side here. There are 1081 choose 36 possible sets (binomial coefficient), so it is out of question to check all of them.
I am not aware of any practicable solution to find the optimal set for the general problem, that is without knowing the 1081x1081 matrix.
For an approximate solution for the general problem, you might want to try a greedy approach, while keeping a history of n sets after each step, with e.g. n = 1000.
So you would start with going through all sets with 2 cards, which is 1081 * 1080 / 2 combinations, look up the value in the matrix for each and pick the n max ones.
In the second step, for each of the n kept sets, go through all possible combinations with a third card (and check for duplicate sets), i.e. checking n * 1079 sets, and keep the n max ones.
In the third step, check n * 1078 sets with a fourth card, and so on, and so forth.
Of course, this won't give you the optimal solution for the general case, but maybe it's good enough for your given situation. You can also take a look at the history, to get a feeling for how often it happens that the best set from step x is caught up by another set in a step y > x. Depending on your matrix, it might not happen that often or even never.
I have a dataframe that with an index of magic card names. The columns are the same index, resulting in a 1081 x 1081 dataframe of each card in my collection paired with each other card in my collection.
I have code that identifies combos of cards that go well together. For example "Whenever you draw a card" pairs well with "Draw a card" cards. I find the junction of those two cards and increase its value by 1. 
Now, I need to find the maximum value for 36 cards.
But, how?
Randomly selecting cards is useless, there are 1.717391336 E+74 potential combinations. I've tried pulling out the lowest values and that reduces the set of potential combinations, but even at 100 cards you're talking about 1.977204582 E+27 potentials. 
This has to have been solved by someone smarter than me - can ya'll point me in the right direction?