Which is the largest positive integer that is 19 times larger than the sum of its digits?
We state that if such a number exists, it can have at most three digits. The sum of the digits of an $n$-digit number $A$ is at most $9 n$, while $A \geq 10^{n-1}$. Therefore, if the condition holds for $A$, then

$$
10^{n-1} \leq A \leq 19 \cdot 9 n < 200 n
$$

which implies

$$
10^{n-3} < 2 n
$$

This is true only for $n=1$, $n=2$, and $n=3$.

For a three-digit number $abc$, our condition means that

$$
a \cdot 10^{2} + b \cdot 10 + c = 19(a + b + c)
$$

After rearranging, we get

$$
81 a - 9 b - 18 c = 0
$$

or

$$
9 a - b - 2 c = 0 \text{. }
$$

The number $abc$ will be exactly 19 times the sum of its digits if and only if the digits satisfy (1). Since $b + 2 c \leq 27$, from (1) we have $9 a \leq 27$, so $a$ is at most 3. If $a=3$, then from (1) we get

$$
b + 2 c = 27
$$

which can only be satisfied if $b=c=9$. Thus, the largest number we get is 399, which satisfies the condition according to (1). The sought number is therefore 399.