The hints in the comment section provides valuable insights to the problem.
I will write a complete answer just to be sure you understand the hints.
There are $3$ cases whereby the sum of the $4$ digits is even,
(1) All $4$ digits are odd.
(2) All $4$ digits are even.
(3) $2$ digits are odd and $2$ digits are even.
First, note that there are $4$ odd digits and $3$ even digits in the set $\{1,2,3,4,5,6,7\}$.
Let's start with Case (1).
Since each digit can only be chosen once, there is $4$ ways of choosing the first odd digit from $\{1,3,5,7\}$,
followed by $3$ ways of choosing the second odd digit from the remaining $3$ digits,
followed by $2$ ways of choosing the third odd digit from the remaining $2$ digits,
followed by $1$ way of choosing the forth odd digit from the remaining $1$ digit.
So, there are $4\times 3 \times 2\times 1 = 4!$ possibilities for Case (1).
Now on to Case (2),
there are $3$ even digits in $\{1,2,3,4,5,6,7\}$, namely $2,4,6$.
Remember that the $4$ digit number must contain different digits, we cannot use either $2$, $4$ or $6$ more than once to construct the $4$ digit number.
And furthermore, there are only $3$ even digits available, so it is impossible to construct any $4$ digit number with no repetition of a digit.
Hence, there is $0$ possibilities for Case (2).
Finally, let's move on to Case (3), which is slightly more difficult.
We first do an unordered selection of $2$ odd digits from $\{1,3,5,7\}$.
which gives $\binom{4}{2}$ possibilities,
followed by an unordered selection of $2$ even digits from $\{2,4,6\}$.
which gives $\binom{3}{2}$ possibilities.
An example would be choosing the digits $1, 5, 2, 4$.
We can quickly see that a valid $4$ digit number would be $1524$.
Note that other permutations of the $4$ digits also form a valid $4$ digit number,
example, $5124$ and $1542$.
and, there are $4!$ permutations in total.
So, by the product rule, there are $\binom{4}{2} \times \binom{3}{2} \times 4!$ possibilities.
Finally, by sum rule, there are a total $4! + 0 +\binom{4}{2} \times \binom{3}{2} \times 4!$ possibilities for the $3$ mutually exclusive cases.
Best Answer
Here's another way . . .
There are$\;5! = 120\;$such numbers.
For each such number $n$, pair it with what I'll call its complement:$\;88888-n$
Thus, there are$\;5!/2 = 60\;$such pairs.
It follows that the sum is just $60\times 88888 = 5333280$.