4. Given some positive integers not exceeding 10 (which can be the same) whose sum is $S$. Find all values of $S$ such that these numbers can definitely be divided into two parts, where the sum of each part does not exceed 70.


4. $S \leqslant 133$.

Obviously, $S \leqslant 140$.
Assume $S \geqslant 134$.
Let $S=134+m(0 \leqslant m \leqslant 6)$.
Consider one number equal to 8, $m$ numbers equal to 10, and $14-m$ numbers equal to 9. At this point, the total sum is $134+m$. Among these 15 numbers, at least eight numbers are in the same group, but the sum of the smallest eight numbers is not less than
$$
1 \times 8 + m \times 9 + (7-m) \times 10 = 78 - m \geqslant 72,
$$
which is a contradiction.

Next, we prove: When $S \leqslant 133$, any numbers with a sum of $S$ that do not exceed 10 can always be grouped according to the problem's requirements.

First, select a group of numbers, let the sum of these numbers be $A$, and satisfy:
(1) $6070$.
If $A \geqslant 63$, then $S-A \leqslant 70$, so the remaining numbers form another group, meeting the problem's requirements.
Consider the case where $A \leqslant 62$.
Let $A=62-x$.
Since any one of the remaining numbers $a$ has $A+a>70$, i.e., $a>8+x$, which means $a \geqslant 9+x$, so,
$x=0$ or 1.
If the remaining numbers do not exceed seven, their total sum does not exceed 70, thus meeting the problem's requirements. Otherwise, the total sum of the remaining numbers $S-A$ is not less than $8 \times (9+x) = 72 + 8x$.
$$
\begin{array}{l}
\text { Hence } S = (S-A) + A \\
\geqslant 72 + 8x + (62 - x) \\
= 134 + 7x > 133.
\end{array}
$$

This contradicts $S \leqslant 133$.
Therefore, the remaining numbers must not exceed seven, and by the previous argument, they must meet the problem's requirements.