14. How many four-digit numbers can be formed using the digits $1$, $9$, $8$, $8$ that leave a remainder of 8 when divided by 11?
Can form 4 numbers that leave a remainder of 8 when divided by 11
14.【Solution】Using $1, 9, 8, 8$, 12 four-digit numbers can be formed, namely $1988, 1898, 1889$, $9188, 9818, 9881, 8198, 8189, 8918, 8981, 8819, 8891$.

Subtracting 8 from them results in 1980, 1890, 1881, 9180, 9810, 9873, 8190, 8181, $8910, 8973, 8811, 8883$.

Among these, only $1980, 1881, 8910, 8811$ are divisible by 11, meaning that using $1, 9, 8, 8$, 4 four-digit numbers can be formed that leave a remainder of 8 when divided by 11, i.e., $1988, 1889, 8918, 8819$.

【Another Solution】What kind of number can be divisible by 11? A rule for determining this is: compare the sum of the digits in the odd positions with the sum of the digits in the even positions. If their difference is divisible by 11, then the given number is divisible by 11; otherwise, it is not.

Now, we need a number that leaves a remainder of 8 when divided by 11. We can consider this: such a number, when added to 3, can be divisible by 11. Therefore, we get the rule for "a number that leaves a remainder of 8 when divided by 11": add the digits in the even positions to get one sum, then add the digits in the odd positions plus 3 to get another sum. If the difference between these two sums is divisible by 11, then the number is a number that leaves a remainder of 8 when divided by 11; otherwise, it is not.

To form a four-digit number that leaves a remainder of 8 when divided by 11 using $1, 9, 8, 8$, we can divide these 4 numbers into two groups, each with 2 digits. One group will be the thousands and tens digits, with their sum denoted as $A$; the other group will be the hundreds and units digits, with their sum plus 3 denoted as $B$. We need to group them appropriately so that the difference can be divisible by 11. Now, there are only 4 ways to group them:
\begin{tabular}{lll} 
& Even & Odd \\
(1) & 1,8 & 9,8 \\
(2) & 1,9 & 8,8 \\
(3) & 9,8 & 1,8 \\
(4) & 8,8 & 1,9
\end{tabular}

After verification, the first (1) grouping method meets the requirement: $A=1+8, B=9+8+3=20, B-A=11$ is divisible by 11. However, the other three groupings do not meet the requirement.

According to the rule, if a number leaves a remainder of 8 when divided by 11, then swapping any two digits in the odd positions or any two digits in the even positions will result in a new number that also leaves a remainder of 8 when divided by 11. Thus, in the first (1) grouping, either 1 or 8 can be the thousands digit, and either 9 or 8 can be the hundreds digit. This results in 4 possible arrangements: 1988, 1889, 8918, 8819.

Answer: 4 numbers can be formed that leave a remainder of 8 when divided by 11.