3. Given several real numbers (which can be the same) belonging to the interval $[0,1]$ whose sum does not exceed $S$. Find the maximum value of $S$ such that these numbers can always be divided into two groups, with the sums of the numbers in each group being $A$ and $B$, respectively, and $A \leqslant 8, B \leqslant 4$.
3. First prove that $S>11.2$ does not meet the problem's requirements.

In fact, when $S>11.2$, take
$$
S=11.2+14 \varepsilon (\varepsilon>0) \text{. }
$$

Assume the given 14 numbers are all equal to $0.8+\varepsilon$. Then, the sum of any ten of these numbers is greater than 8, and the sum of any five of these numbers is greater than 4. Therefore, the group with sum $A$ can have at most nine numbers, and the group with sum $B$ can have at most four numbers.
Thus, when $S>11.2$, it does not meet the problem's requirements.
Next, prove that when $S \leqslant 11.2$, for any $S$, the problem's requirements are satisfied.

Consider some of the given numbers whose sum is $S_{1}$, and $78 \Rightarrow b>0.8+x . \\
\text{ By } A+5(0.8+x)=11.2+4 x>S,
\end{array}
$$

we know that at most four numbers remain.
Since each number is no more than 1, then $B \leqslant 4$, which meets the problem's conditions.