Solution:

The maximum sum that Barbara is sure to collect is 32 euros.

At each move, Barbara can at least double the minimum distance between the remaining numbers. Indeed, on the first move she can remove all the odd numbers, and on the subsequent ones, regardless of Alberto's moves, she can remove the $2^{\circ}$, the $4^{\circ}$, the $6^{\circ}$, ... of the remaining numbers according to increasing order. After her 5 moves, the minimum distance between the remaining numbers will be at least $2^{5} = 32$.

On the other hand, Alberto can at each move at least halve the maximum distance between the remaining numbers. Indeed, after Barbara's first move there will remain $2^{10} + 1 - 2^{9} = 2^{9} + 1$ numbers; among these, necessarily either among those less than $2^{9}$ or among those greater than $2^{9}$ there will not be more than $2^{8}$. Alberto can therefore remove those remaining, arranging things so that only numbers remain in the interval $[0, 2^{9}]$ or only numbers in the interval $[2^{9}, 2^{10}]$.

Before Alberto's next move, $2^{9} + 2^{8} + 2^{7}$ numbers will already have been removed, and so there will remain $2^{7} + 1$ numbers, contained in an interval of length $2^{9}$. Consequently, either in the first half or in the second half of the remaining interval (excluding the central number), there will not remain more than $2^{6}$, and so Alberto will be able to remove them all, leaving only numbers in an interval of length $2^{8}$.

Continuing with this strategy up to his fifth and last move, Alberto will be able to arrange things so as to leave only two numbers in an interval of length $2^{5} = 32$, and hence with distance not exceeding 32.

In conclusion, Barbara is sure to earn at least 32 euros, Alberto is sure not to have to pay out more than 32 euros, and therefore 32 euros is the maximum sum that Barbara is sure to be able to collect.