To write the problem in somewhat mathematical terms, we have three digits $d_1 \in \{1, \dots, 9\}$, $d_2,d_3 \in \{0, 1, \dots, 9\}$ that form a $3$-digit number
$$x = d_1 \cdot 10^2 + d_2 \cdot 10 + d_3$$
Now, we want to find all possible values of $x$ such that at least one of the following holds:
- $d_1 + d_2 = d_3$,
- $d_1 + d_3 = d_2$,
- $d_2 + d_3 = d_1$
Let's start with the first condition. We may pick $d_1$ freely, which gives us $9$ options. However, since $d_3 \in \{0, 1, \dots, 9\}$, we must always choose $d_2 \in \{0, \dots, 9-d_1\}$. Hence:
$$\# \{ x \mid d_1 + d_2 = d_3\} = \sum_{d_i = 1}^{9}(10-d_i) = 45$$
The same reasoning holds for the second condition:
$$\# \{ x \mid d_1 + d_3 = d_2\} = \sum_{d_i = 1}^{9}(10-d_i) = 45$$
However, the last condition is different! We pick $d_2$ freely which gives us $10$ options, and we must choose $d_3 \in \{0, \dots, 9-d_2\}$. However, we also have to exclude one case: $d_2 = d_3 = 0$, because then $d_1$ would also be $0$. Hence:
$$\# \{ x \mid d_2 + d_3 = d_1\} = \sum_{d_i = 0}^{9}(10-d_i)-1 = 54$$
We now have $45 + 45 + 54 = 144$ numbers $x$ such that either of the conditions hold. However, some numbers are counted twice. It is easy to see that a number is counted twice if and only if $d_i = d_j$ and $d_k = 0$ for $i,j,k$ being some permutation of $1,2,3$. Since $d_1 \neq 0$, the multiples we need to discard are those of the form $d_2 = 0$ or $d_3 = 0$. There are $9$ of each ($d0d$ and $dd0$ with $d = 1, \dots, 9$), so we remove $18$ multiples. There is no number for which all $3$ conditions hold, since that would require $d_1 = d_2 = d_3 = 0$.
The solution is then:
$$\#\{x\} = 144- 18 = 126.$$
Let $S$ be the set of numbers where at least one digit is in the range $\{1,\dots,8\}$. I claim that exactly one fourth of the numbers in $S$ have a digit sum which is a multiple of $4$. To do this, we will partition $S$ into groups of four, where the numbers in each group have different remainders modulo $4$.
Let $S_1$ be the set of numbers in $S$ whose first digit is in $\{1,\dots,8\}$, and let $x$ be some number in $S_1$. If the first digit of $x$ is in $\{1,2,3,4\}$, the group containing $x$ is made by changing that digit to the other numbers in $\{1,2,3,4\}$. For example, the group containing $34682$ would be $\{14682, 24682, 34682, 44682\}$. You do the corresponding thing if the first digit is in $\{5,6,7,8\}$.
But what do we do if the first digit is $0$ or $9$? You then consider $S_2$, the set of numbers whose first digit is $0$ or $9$, and whose second digit is in $\{1,\dots,8\}$, and do the same procedure in the first bullet to the second digit.
Then, you let $S_3$ be the set of numbers each of whose first two digits are $0$ or $9$, and whose third digit is in $\{1,\dots,8\}$, and do the same thing. Same with $S_4$.
Since $S$ is the non-overlapping union of $S_1,S_2,S_3$ and $S_4$, we have partitioned $S$ into groups of four as desired. Therefore, we need to count the number of numbers in $S$ and divide by $4$ to find the number of desired numbers in $S$. To count $S$, use complementary counting.
What about the numbers outside $S$? There are only $16$ of these, and you can count the number of these whose digit sum is a multiple of four directly.
Best Answer
If we exclude $0$ and remove the limit on the number of digits appearing we can find a nice stars and bars approach to get the number. If the final digit is $9$ we start with $1111111119$. There are eight potential spaces for bars between the $1$s and after choosing which to receive bars you sum up the $1$s between them, so $11|111|11119$ would give us $2349$. As there are eight spaces there are $2^8=256$ numbers ending in $9$. Similarly, if the last digit is $n$ there are $2^{n-1}$ numbers, so the total number is $\sum_{i=1}^9 2^{i-1}=2^9-1=511$. This is the same approach suggested by lulu's comment.