Let us consider all three-digit numbers whose digits are all different and nonzero. Each such number can be written as $100a + 10b + c$, where $a$, $b$, $c$ are digits from $1$ to $9$, all different.

The total number of such numbers is $9 \times 8 \times 7 = 504$.

Let $S$ be the sum of all such numbers.

Let us fix the digits $a$, $b$, $c$ in all possible orders. For each digit, it appears in each position (hundreds, tens, units) the same number of times, because the digits are permuted among the three positions.

For each digit $d$ ($1 \leq d \leq 9$), the number of ways to choose the other two digits is $8 \times 7 = 56$, and for each such choice, $d$ appears in each position exactly $2! = 2$ times (since there are $3! = 6$ permutations, and for each position, $2$ permutations place $d$ there).

So, for each digit $d$, it appears $56 \times 2 = 112$ times in each position.

Therefore, the total sum is:

$$S = \sum_{d=1}^9 112 \cdot (100d + 10d + d) = \sum_{d=1}^9 112 \cdot 111d = 112 \cdot 111 \cdot \sum_{d=1}^9 d$$

But $\sum_{d=1}^9 d = 45$.

So,
$$S = 112 \cdot 111 \cdot 45$$

Let us factor this number:

$112 = 16 \times 7 = 2^4 \times 7$

$111 = 3 \times 37$

$45 = 5 \times 9 = 5 \times 3^2$

So,
$$S = (2^4 \times 7) \times (3 \times 37) \times (5 \times 3^2)$$

Multiply the $3$'s:

$3 \times 3^2 = 3^3$

So,
$$S = 2^4 \times 3^3 \times 5 \times 7 \times 37$$

Thus, the prime divisors of $S$ are $2$, $3$, $5$, $7$, and $37$.

Therefore, the sum $S$ has at least three different prime divisors (in fact, it has five).