1. Determine the number of pairs $(a, b)$ of natural numbers $(1 \leqq a<b \leqq 86)$ for which the product $a b$ is divisible by three.
1. First, we calculate the total number of pairs of numbers such that \(1 \leqq ab\). The number of all pairs \((a, b)\) of natural numbers such that \(1 \leqq a < b \leqq 86\) is therefore equal to \(\frac{1}{2} \cdot 86 \cdot 85 = 3655\). (This is also the number of all unordered pairs of natural numbers from the set C, which is the combination number \(\left(\begin{array}{c}86 \\ 2\end{array}\right) = 3655\)).

The product \(ab\) is divisible by three if and only if at least one of the factors \(a, b\) is divisible by three. Since there are exactly 28 numbers in the set C that are divisible by three, there are \(86 - 28 = 58\) numbers in C that are not divisible by three. Therefore, we can form \(\frac{1}{2} \cdot 58 \cdot 57 = 1653\) pairs of different natural numbers \((a, b)\) such that \(1 \leqq a < b \leqq 86\), and the product \(ab\) is not divisible by three.

The number of all pairs \((a, b)\) of natural numbers \((1 \leqq a < b \leqq 86)\) for which the product \(ab\) is divisible by three is equal to \(3655 - 1653 = 2002\).

Another solution. Let A and B be the sets of all pairs \((a, b) (1 \leqq a < b \leqq 86)\) in which the number \(a\), respectively \(b\), is divisible by three. Among the numbers in the set \(C = \{1, 2, \ldots, 86\}\), there are 28 numbers divisible by three (these are the numbers \(3, 6, 9, \ldots, 84\)). For each number \(a \in C\), there are \(86 - a\) numbers \(b \in C\) such that \(a < b\). Therefore, the number of all elements of the set A is equal to

\[
\begin{aligned}
(86 - & 3) + (86 - 6) + (86 - 9) + \cdots + (86 - 84) = \\
& = 28 \cdot 86 - 3 \cdot (1 + 2 + 3 + \cdots + 28) = \\
& = 28 \cdot 86 - 3 \cdot \frac{1}{2}((1 + 28) + (2 + 27) + (3 + 26) + \cdots + (28 + 1)) = \\
& = 28 \cdot 86 - 3 \cdot \frac{1}{2} \cdot 29 \cdot 28 = 2408 - 1218 = 1190.
\end{aligned}
\]

For each number \(b \in C\), there are \(b - 1\) numbers \(a \in C\) such that \(a < b\). Therefore, the number of all elements of the set B is equal to

\[
\begin{aligned}
(3 - 1) + (6 - 1) + (9 - 1) + \cdots + (84 - 1) & = 3 \cdot (1 + 2 + \cdots + 28) - 28 = \\
& = 1218 - 28 = 1190.
\end{aligned}
\]

The intersection of sets A and B contains pairs of numbers \((a, b)\) in which both components \(a\) and \(b\) are divisible by three, and \(a < b\). According to the reasoning from the initial solution, there are \(\frac{1}{2} \cdot 28 \cdot 27 = 378\) such pairs. The number of elements in the union of sets A and B, i.e., the number of all pairs \((a, b)\) of natural numbers \((1 \leqq a < b \leqq 86)\) for which the product \(ab\) is divisible by three, is equal to the sum of the elements of sets A and B minus the number of elements of their intersection, i.e., \(1190 + 1190 - 378 = 2002\). Full solution is worth 6 points.