The Gaussian elimination in this case will work as follows.
First sort the numbers in a non-ascending order. Now visualize the binary representations of all the numbers in a matrix. Consider them to be non-negative 64 bit integers.
(Note: you don't have to actually convert them to binary or make the matrix, use bit operations on an array of numbers)
Then carry out the "Forward Elimination" phase of Gaussian Elimination. First find the MSB position (i.e. the first 1) of the very first number. Make sure that all the numbers below it don't have have a '1' at this position. If they do then just XOR them with the number at top and their bits at this position will become zero.
After that move on to the next row and next column, and repeat the process. If at anytime you find that the row you are working on does not have a 1 at the desired position, then carry out "Pivoting". Look for any number below that may have a 1 at this position and swap the rows i.e. the numbers. If all numbers below also have 0's then stay at the same row and move on to the next column, and keep repeating the process if all you find are 0's.
(For this problem you don't have to be at the last row and last column at the same time.)
Once elimination is done simply traverse each row of the matrix. Have a variable called result which is initially 0. You have to take the first row since it has the highest MSB. By taking I mean carrying out XOR operation with your current result variable. Then move on to the next row and column. If you see that your result has a 0 at this column then XOR with the number at this row since it can only increase your sum or keep it the same.
In this way check all the rows, and the result you finally have is the maximum xor sum.
(An easier way to think of this would be that you only XOR a row with your current result if it increases the current result.)
What we are doing here is trying to make each number 1 bit shorter than the previous number. This makes it easier for us to think of the XOR operations without thinking of how it may effect the bits to the left. You may have to think a little to understand why this solution is valid.