Your set of passwords can be decomposed into three disjoint sets:
1) $2k,*, *, *, *, 2l$
2) $2k,*, *, *, *, 2l+1$
3) $2k+1,*, *, *, *, 2l+1$
In each one of these sets, there are
$$
5\cdot10^4\cdot5
$$
passwords.
We first count the number of ways to produce an even number. The last digit can be any of $2$, $4$, or $6$. So the last digit can be chosen in $3$ ways.
For each such choice, the first digit can be chosen in $6$ ways. So there are $(3)(6)$ ways to choose the last digit, and then the first.
For each of these $(3)(6)$ ways, there are $5$ ways to choose the second digit. So there are $(3)(6)(5)$ ways to choose the last, then the first, then the second.
Finally, for each of these $(3)(6)(5)$ ways, there are $4$ ways to choose the third digit, for a total of $(3)(6)(5)(4)$.
Similar reasoning shows that there are $(4)(6)(5)(4)$ odd numbers. Or else we can subtract the number of evens from $840$ to get the number of odds.
Another way: (that I like less). There are $3$ ways to choose the last digit. Once we have chosen this, there are $6$ digits left. We must choose a $3$-digit number, with all digits distinct and chosen from these $6$, to put in front of the chosen last digit. This can be done in $P(6,3)$ ways, for a total of $(3)P(6,3)$.
Best Answer
Hint: There are two types of $4$ digit odd integers with all digits different: (i) First digit is even ($2,4,6,8$) or (ii) first digit is odd. Count the Type (i) numbers, the Type (ii) numbers, and add.
Type (i): There are $4$ choices for the first digit. For each of these choices there are $5$ choices for the last digit. For each of the choices we have made so far, there are $8$ choices for the second digit, and for each such choice $7$ choices for the third, a total of $(4)(5)(8)(7)$.
Now it's your turn: make a similar calculation for Type (ii).
The analysis for the even numbers will be roughly similar. The two cases are (i) last digit is $0$ and (ii) last digit is not $0$.
Remark: The tricky thing is that the first digit cannot be $0$. A slightly nicer approach, I think, is to first allow $0$ as a first digit, and then take away the "numbers" with first digit $0$, which we should not have counted.
So for odd numbers, if we allow $0$ as a first digit, we can choose the last digit in $5$ ways, and then for the rest we have $(9)(8)(7)$ choices, for a total of $(5)(9)(8)(7)$.
Now if we have $0$ as the first digit, the rest can be filled in $(5)(8)(7)$ ways.
So our answer is $(5)(9)(8)(7)-(5)(8)(7)=2240$.
For the evens, the same analysis gives $(5)(9)(8)(7)-(4)(8)(7)$.