11.5. On each of ten cards, a real number is written. For each non-empty set of these cards, the sum of all numbers written on the cards in this set was found. It is known that not all of the obtained sums are integers. What is the maximum possible number of integer sums that could have resulted?
Answer: 511.

Solution: It is clear that at least one of the cards has a non-integer number, otherwise the sum of the numbers in any set would be an integer. We select one of these cards and call it $a$, and the number on it $-x$. For each set $C$ that does not contain $a$ (including the empty set), we associate the set $C \cup\{a\}$. It is clear that different sets will be mapped to different sets, because if $C \cup\{a\}$ and $D \cup\{a\}$ coincide, then $C$ and $D$ coincide. Therefore, the constructed different pairs do not have common elements.

Notice that any set $C$ will end up in one of these pairs. If it does not contain $\{a\}$, it will fall into a pair by construction; if it does contain $\{a\}$, it is paired with the set from $C$ without $\{a\}$.

Thus, all sets are divided into pairs of the form ( $C, C \cup\{a\}$ ). We discard the pair ( $\varnothing, a$ ) (since neither of these sets gives an integer sum) and look at the remaining ones. The difference in the sums of the numbers on the cards in the sets of a pair differs by the number $x$, so in each pair there is no more than one integer. Since there are $\frac{2^{10}-2}{2}=2^{9}-1=511$ pairs, there are no more than 511 sets with an integer sum.

If our set contains 9 cards with 0 and one with the number $\frac{1}{2}$, then the sum of the numbers on the cards of $2^{9}-1=511$ sets will be 0, which is an integer.

Comment: Only the answer - 1 point.

The answer is given and an example is provided where 511 integer sums are obtained - 2 points.

The solution includes the idea of dividing sets into such pairs that no more than one set can be taken from each - 1 point.

It is proven that there will be no more than 511 integer sums - 5 points.

Remark: 511 integer sums can also be obtained in the case where no integer is written on any of the cards. For example, if $\frac{1}{2}$ is written on each of them, we will get exactly 511 integer sums.