Four, (50 points) Given that there are several numbers (which can be the same) in the interval $[0,1]$, and their sum does not exceed $S$. Find the maximum value of $S$ such that these numbers can always be divided into two groups, with the sum of each group not exceeding 11.
Four, the maximum value of $S$ is $\frac{253}{12}$.
First, prove: $S \leqslant \frac{253}{12}$.
If $S>\frac{253}{12}$, then take
$$
\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{23} \\
=\frac{11}{12}+\frac{\varepsilon}{23}\left(\varepsilon=S-\frac{253}{12}>0\right) .
\end{array}
$$

By the pigeonhole principle, there must be a group of more than 11 numbers, with a total sum not less than $11+\frac{12}{23} \varepsilon$, which is a contradiction.
Next, prove: $S=\frac{253}{12}$ satisfies the condition.
Let these several numbers be arranged in non-decreasing order as $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{t}$, and let $S_{i}=\sum_{j=1}^{i} x_{j}$.
Let the maximum value of the elements in the set $\left\{i \mid S_{i} \leqslant 11\right\}$ be $i_{0}$.
If $S_{i_{0}} \geqslant \frac{121}{12}$, then $S-S_{i_{0}} \leqslant \frac{253}{12}-\frac{121}{12}=11$, indicating that the numbers have already been divided into two groups, each with a sum not exceeding 11;
If $S_{i_{0}} < \frac{121}{12}$, then by the definition of $i_{0}$, we know $\delta S,
$$

which is a contradiction.
Thus, $t-i_{0} \leqslant 11$.
Therefore, $S-S_{i_{0}} \leqslant\left(t-i_{0}\right) x_{t} \leqslant t-i_{0} \leqslant 11$.
This shows that the numbers have already been divided into two groups, each with a sum not exceeding 11.
In conclusion, the maximum value of $S$ is $\frac{253}{12}$.