1. Determine the number of all four-digit natural numbers that are divisible by six and in whose notation exactly two ones appear.
1. For a number to be divisible by six, it must be even and have a digit sum divisible by three. Let's denote $b$ as the digit in the units place (which must be even, $b \in\{0,2,4,6,8\}$) and $a$ as the digit that, together with the digits $1,1 (a \neq 1)$, is on the first three places of a four-digit number that meets the requirements of the problem.

For the sum of the digits $a+1+1+b$ of such a number to be divisible by three, the number $a+b$ must give a remainder of 1 when divided by three. For $b \in\{0,6\}$, we have the options for $a$ as $a \in\{4,7\}(a \neq 1)$, for $b \in\{2,8\}$, $a \in\{2,5,8\}$, and finally for $b=4$, $a \in\{0,3,6,9\}$. For each chosen $b$ and corresponding $a \neq 0$, there are clearly three ways to arrange the digits 1,1 and $a$ in the first three places, which is a total of $(2 \cdot 2+2 \cdot 3+3) \cdot 3=39$ possibilities. For $a=0$ (when $b=4$), there are only two possibilities (the digit zero cannot be the first digit of a four-digit number).

In total, there are 41 four-digit natural numbers that meet the conditions of the problem.

For a complete solution, award 6 points, of which 3 points are for using the insight that the last digit must be even. If, however, someone analyzes all 81 remaining cases of the two digits and correctly determines the number of suitable numbers, they also deserve 6 points. When describing an efficient method, do not deduct more than 2 points for any arithmetic errors in determining the final count.