274. How many natural numbers divisible by 3 or 7 are there among the natural numbers from 1 to 500?
$\triangle$ First, let's calculate how many natural numbers from 1 to 500 are divisible by 3. Let's list all such numbers:

$$
3, \quad 6=3 \cdot 2, \quad 9=3 \cdot 3, \quad 12=3 \cdot 4, \ldots, \quad 498=3 \cdot 166
$$

From this, we can see that the number of such numbers is 166.

Now let's find how many natural numbers within the same range are divisible by 7. Let's list them:

$$
7, \quad 14=7 \cdot 2, \quad 21=7 \cdot 3, \ldots, \quad 497=7 \cdot 71
$$

Thus, the number of these numbers is 71.

But this is not all: among these numbers, there are some that are divisible by both 3 and 7, which means they are also divisible by 21. Since these numbers are included in both the 166 numbers divisible by 3 and the 71 numbers divisible by 7, they are counted twice in \(166+71=237\) numbers. Let's count how many there are. Let's list the numbers from 1 to 500 that are divisible by 21:

$$
21, \quad 42=21 \cdot 2, \quad 63=21 \cdot 3, \ldots, \quad 481=21 \cdot 23
$$

Thus, there are 23 such numbers. Therefore, the total number of numbers divisible by 3 or 7 is \(237-23=214\).

Answer: 214.

When solving this problem, the principle of inclusion and exclusion for the case of two sets can be applied:

$$
n(A \cup B)=n(A)+n(B)-n(A \cap B)
$$

(see [18], § 4, formula (1)). Let's also recall the similar formula for the case of three sets:

$$
\begin{gathered}
n(A \cup B \cup C)= \\
=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C) .
\end{gathered}
$$