jmc

Case 1: The last two digits of our integer are equal. There are 10 possibilities for these last two digits and 4 choices for the hundred's digit, of a total of 40 possibilities. (Note that this case includes 111, 222, 333, and 444.)

Case 2: The first two digits are equal and the third is different. This occurs in $4\cdot 9=36$ ways, since we can choose the repeated digit in 4 ways and the remaining digit in 9.

Case 3: The first and third digits are equal while the second is different. This also occurs in 36 ways.

Thus we have a total of $40+36+36=\boxed{112}$ integers.

OR

Another way to solve this problem is to find how many three-digit integers less than 500 have no digits that are the same. The first digit must be 1, 2, 3, or 4. The second digit can be any of the 9 digits not yet chosen, and the third digit can be any of the 8 digits not yet chosen, so there are a total of $4\cdot 9\cdot 8=288$ three-digit integers that have no digits that are the same and are less than 500. There are a total of $500-100=400$ three-digit integers that are less than 500, so we have a total of $400-288=\boxed{112}$ integers that fit the problem. (Solution by Alcumus user chenhsi.)