4. How many sets of three consecutive three-digit natural numbers have a product that is divisible by 120?
First method:

The prime factorization of the number 120 is $120=2^{3} \cdot 3 \cdot 5$.

Therefore, for a number to be divisible by 120, it must be divisible by 3, 5, and $8. \quad 1$ POINT Among three consecutive natural numbers, exactly one is divisible by 3, so the product of three consecutive natural numbers is always divisible by 3.

1 POINT Let the first of the three consecutive numbers be $m$ and even.

Then the third number $(m+2)$ is also even. One of the two consecutive even numbers is divisible by 2, and the other by 4, so their product is divisible by 8.

1 POINT

Divisibility by 5 will be satisfied if any of the three numbers is divisible by 5. This is possible if and only if the first or third (even) number ends in 0, or if the second (odd) number ends in 5.

In the desired triples, the first number can be of the form $\overline{a b 0}$, the third number can be of the form $\overline{c d 0}$, or the second number in the triple can be of the form $\overline{e f} 5$.

1 POINT

Thus, in the desired triples, the first number can be of the form $\overline{a b 0}$, where the digits $a$ and $b$ are any digits from the set $\{1,2,3,4,5,6,7,8,9,0\}$ with the digit $a \neq 0$, and there are $\mathbf{9 0}$ such numbers.

Similarly, the third number can be of the form $\overline{c d 0}$. There are also 90 possibilities, but in the first triple $(98,99,100)$, two-digit numbers appear. There are 89 such triples.
Thus, the number of triples where the first or third number ends in 0 is $90+89=179$. ..... 1 POINT
The second number in the triple can be of the form $\overline{e f 5}$. There are $\mathbf{9 0}$ such triples. ..... 1 POINT
Now let the first number $m$ in the triple be odd.
Since the third number $(m+2)$ is also odd, it is necessary that the second number $(m+1)$ be divisible by 8. $\quad 1$ POINT
Since we are combining divisibility by 8 and divisibility by 5, it is sufficient to consider the first 40 possibilities.
The middle number $(m+1)$ can be 104, 112, 120, 128, or 136, but either it or its immediate predecessor or successor must be divisible by 5.
This happens in three cases: (103, 104, 105), (119, 120, 121), and (135, 136, 137). ..... 1 POINT
There are 900 three-digit numbers, and from $900=40 \cdot 22+20$ it follows that the desired numbers are $22 \cdot$ plus the number 984 in the last 20 numbers. There are 67 such triples. ..... 1 POINT
The total number of triples is $90+89+90+67=336$. ..... $1 \mathrm{POINT}$
TOTAL 10 POINTS

Note 1: The last point is awarded only if the result is correct.

Note 2: Listing a certain number of valid triples (of course not all) that satisfy the conditions of the problem without providing a general strategy for counting all such triples should be evaluated with 0 POINTS.