I think considering the two different cases you mentioned separately is best. To avoid further case division, I'd proceed like this:
Case with leading digit 4: since an even digit has to go in the rightmost position, there are $5 \choose 3$ ways to choose the positions of the 3 odd numbers amongst the other 5 positions, then 3 ways to order them, and $3!$ ways to order the even numbers, for 180 total.
Case with leading digit 5: similarly, there are $5 \choose 2$ ways to choose the positions of the two odd digits (both 3's), only 1 way to order them, then $\frac{4!}{2}$ ways to order the even digits (dividing by 2 since there are 2 indistinguishable 4's), for 120 total.
So all up there are 300 such numbers.
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:
$$\#\{x\} = 144- 18 = 126.$$
Best Answer
HINT: The possible sums of two digits are the integers from $0$ through $18$. There’s just one way to get a sum of $0$ or $18$. There are two ways to get a sum of $1$, $01$ and $10$, and two ways to get a sum of $17$, $89$ and $98$. It’s easy enough to work out the number of ways to get each possible sum:
$$\begin{array}{r|rr} \text{Sum}&0&1&2&3&4&5&6&7&8&9\\ \text{Sum}&18&17&16&15&14&13&12&11&10\\ \hline \text{Nr. of ways}&1&2&3&4&5&6&7&8&9&10 \end{array}$$
To get a number whose first two digits sum to $6$, say, and whose last two digits also sum to six, you must combine one of the $7$ possible pairs for the first two digits with one of the same $7$ possible pairs for the last two digits. You can do that in $7^2=49$ ways.
Can you finish the calculation from there?