The sum of several (not necessarily different) positive integers not exceeding $10$ is equal to $S$.
Find all possible values of $S$ such that these numbers can always be partitioned into two groups with the sum of the numbers in each group not exceeding $70$.

[i](I. Voronovich)[/i]
1. **Understanding the Problem:**
   We need to find all possible values of \( S \) such that any set of positive integers not exceeding 10, whose sum is \( S \), can be partitioned into two groups with each group's sum not exceeding 70.

2. **Initial Observations:**
   - If \( S \) does not satisfy the property, then \( S+1 \) also does not satisfy the property because we can add a 1 to the set.
   - Therefore, it suffices to show that 133 satisfies the property and 134 does not.

3. **Showing \( S = 134 \) does not work:**
   - Consider the set \( \{9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8\} \).
   - The sum of this set is \( 9 \times 14 + 8 = 126 + 8 = 134 \).
   - Any partition of this set must have at least one group with at least 8 numbers.
   - The sum of the smallest 8 numbers (all 9s) is \( 9 \times 8 = 72 \), which exceeds 70.
   - Hence, \( S = 134 \) does not satisfy the property.

4. **Showing \( S = 133 \) works:**
   - We need to show that for any set \( T \) of positive integers not exceeding 10, there exists a subset \( A \subseteq T \) with the sum of its elements in the interval \([63, 70]\).

5. **Case 1: There is an element \( t \) in \( T \) which is in \([2, 8]\):**
   - Start by adding \( t \) to \( A \).
   - Continually add random elements of \( T \) to \( A \) until the sum is at least 63.
   - If the sum of the elements in \( A \) is in \([63, 70]\), we are done.
   - Otherwise, if the sum is in \([71, 72]\), we can delete \( t \) from \( A \) to bring the sum back to \([63, 70]\).

6. **Case 2: All elements of \( T \) are either 1, 9, or 10:**
   - If there are at least two 1s, combine them into a 2 and apply the logic from Case 1.
   - Suppose there is at most one 1 in \( T \).
   - The sum of the other elements is at least 132, implying there are at least 7 elements which are either 9 or 10.
   - The sum of these 7 elements lies in the interval \([63, 70]\).

7. **Conclusion:**
   - Since we have shown that \( S = 133 \) works and \( S = 134 \) does not, the possible values of \( S \) are all \( S \leq 133 \).

The final answer is \( \boxed{ S \leq 133 } \).