All of the digits of a three-digit integer are distinct and non-zero. Furthermore, the three-digit integer is divisible by 19. Find the largest three-digit integer that has these properties.

We first list out possible three-digit multiples of $19$. Because the remainder of $100\div19$ is $5$, the digits of multiples of $19$ must be between $055$ and $094$ inclusive. After manually checking divisibility, we find that the only three-digit numbers that fit this condition are $095, 114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 304, 323, 342, 361, 380$. Notice that if we reverse the order of each of these numbers, we go from a relatively small number to relatively an even greater number, so we try reversing $\boxed{380}$, which satisfies all the given conditions.

To find the largest three-digit integer with distinct and non-zero digits that is divisible by 19, we can start with the largest possible three-digit number and work backwards.

The largest three-digit number is 999, but this number is not divisible by 19. So, we need to find the next largest three-digit number.

To find this number, we can begin by listing the three-digit numbers in descending order:

998, 997, 996, 995, ..., 100

We then check each of these numbers to see if they are divisible by 19.

Starting with 998, we can divide it by 19 to see if it is divisible evenly:

998 ÷ 19 = 52 remainder 10

Since there is a remainder of 10, 998 is not divisible by 19. We continue to check the remaining numbers.

997 ÷ 19 = 52 remainder 9

996 ÷ 19 = 52 remainder 8

995 ÷ 19 = 52 remainder 7

994 ÷ 19 = 52 remainder 6

993 ÷ 19 = 52 remainder 5

992 ÷ 19 = 52 remainder 4

991 ÷ 19 = 52 remainder 3

990 ÷ 19 = 52 remainder 2

989 ÷ 19 = 52 remainder 1

988 ÷ 19 = 52 remainder 0

We have found a number, 988, that is divisible by 19. However, this number does not have distinct digits.

Continuing our search, we can try the next largest number, 987, and check if it is divisible by 19.

987 ÷ 19 = 51 remainder 18

Since there is a remainder of 18, 987 is not divisible by 19.

We continue this process, checking each number until we find a three-digit number that is divisible by 19.

986 ÷ 19 = 51 remainder 17

985 ÷ 19 = 51 remainder 16

984 ÷ 19 = 51 remainder 15

983 ÷ 19 = 51 remainder 14

982 ÷ 19 = 51 remainder 13

981 ÷ 19 = 51 remainder 12

980 ÷ 19 = 51 remainder 11

979 ÷ 19 = 51 remainder 10

978 ÷ 19 = 51 remainder 9

977 ÷ 19 = 51 remainder 8

976 ÷ 19 = 51 remainder 7

975 ÷ 19 = 51 remainder 6

974 ÷ 19 = 51 remainder 5

973 ÷ 19 = 51 remainder 4

972 ÷ 19 = 51 remainder 3

971 ÷ 19 = 51 remainder 2

970 ÷ 19 = 51 remainder 1

969 ÷ 19 = 51 remainder 0

We have found a number, 969, that is divisible by 19 and has distinct digits.

Therefore, the largest three-digit integer with distinct and non-zero digits that is divisible by 19 is 969.