A numbers problems

I could be wrong, but I think the answer is 165. Let's start with the 1000's first. For any number that starts with one and ends in two, the center two numbers must add to 2-1. Since the two center numbers have to add to 1 and the number starts with 1 and ends in 2, we get 1102 and 1012. It's similar to a number that starts with 1 and ends in a 3. The center two numbers must add to 3-1. There are 3 different ways 2 numbers can add to 2: 0,2 1,1 and 2,0. Therefore, we now have 1023, 1113, and 1203. Since there is 1 way to make a number that ends in 1, 2 ways that end in 2o, 3 ways for 3 etc, we can continue this pattern for 4 through 9, making a total of 1+2+3+4+5...+9, 45. So there are 45 four digit numbers that start with 1. For numbers that start with 2, we'll start with a number that starts with 2 and ends in 1, but 1-2 = -1, and we can't have two positive numbers that add to a negative number, so no number that starts with 2 and ends in 1 works. For numbers that start with 2 and end in 2, only 2002 works. For numbers that start with 2 and end in 3, we get two possible combinations, 0,1 and 1,0. Notice how as we increase our starting number the possibilities for each end number decrease by one from what the previous starting number is. Numbers that start with 1 had 1 combination for 1, 2 for 2,3 for 3, etc. For numbers that start with 2, we get 1 possibility for 2, 2 for 3, 3 for 4, etc. Since there are 9 numbers we can end with for starting with 1, there will be 8 ending numbers for 2, 7 for 3, etc. Since we loose 1 possibility each time we increase the starting number by 1, there will be 9 fewer combinations for numbers that start with 2, or 35. There will be 8 fewer numbers that start with 3 than 2, 7 fewer that start with 4 than start with 3 etc. So for all numbers 1000-9999, we started with 45 for 1, 45-9 for 2, 45-9-8 for 3, 45-9-8-7 for 4 and so on. 45+36+28+21+15+10+6+3+1 = 165. So (I think) there are 165 numbers from 1000-9999 that follow your rule.