1. How many four-digit natural numbers can be written using the digits $0,1,2,3,4,5,6$, if all digits must be different and the sum of the digits must be $15?$
1. Note that the sum of all allowed digits is $0+1+2+3+4+5+6=$ $\frac{6 \cdot 7}{2}=21$. Since we need to form a four-digit number from some of these digits whose sum must be 15, we need to discard three digits whose sum is 6. Clearly, they can only be one of the following: $\{0,1,5\},\{0,2,4\}$ or $\{1,2,3\}$.

In the first case, we can use the digits $2,3,4,6$, and they can be arranged in $4!=24$ ways. In the second case, we can use the digits $1,3,5,6$, and they can be arranged in $4!=24$ ways. In the third case, we can use the digits $0,4,5,6$, and when arranging them, we must ensure that 0 is not in the first position; thus, for the leftmost digit, we choose one of $4,5,6$, and then the remaining three can be arranged in $3!=6$ ways, so in this case, we have a total of $3 \cdot 6=18$ numbers.

Therefore, the solution to the problem is $24+24+18=66$.