Let's determine those positive integers which are 14 times as large as the sum of their digits.
Solution. Let the number we are looking for be denoted by $A$, and the sum of its digits by $x$. According to the problem's requirement, $A=14 x$. We know that if we subtract the sum of the digits of a number from the number itself, we get a number divisible by 9; thus $14 x-x=13 x$ is divisible by 9. Therefore, $x$ must also be divisible by 9. Since the number we are looking for is also divisible by 14 (which is relatively prime to 9), it must necessarily be divisible by $9 \cdot 14=126$. Among the three-digit multiples of 126, the first four are: $126, 2 \cdot 126=252, 3 \cdot 126=378$, $4 \cdot 126=504$. Among these, only 126 is a solution. There is no other solution among the three-digit numbers, since the sum of the digits of such a number is less than $3 \cdot 10=30$, and 14 times the sum of the digits is less than $30 \cdot 14=420$.

We will show that there is no solution to the problem among the four- or more-digit numbers. If $A$ were such an $n$-digit number, where $n$ is at least 4, then $A$ is at least $10^{n-1}$, while the sum of its digits is at most $9 n$. Therefore, $A=14 x \leq 14 \cdot 9 n=126 n$, so $10^{n-1} \leq 126 n$. However, for $n=4$, $126 n=126 \cdot 4=504<1000=10^3$, and if the inequality holds for some $n$, then for the next value of $n$:

$$
10^{n}=10^{n-1} \cdot 10>126 n \cdot 10>126 n \cdot \frac{n+1}{n}=126(n+1)
$$

so the inequality also holds for $n+1$; thus it holds for all $n \geq 4$. The only number that meets the conditions of the problem is 126.