What I can't seem to get over, is that there are an equal number of odd and even numbers, so, why is the answer not 1/2 ?
Because they are not complementary events. There is another possibility.
The probability that both dice show even numbers is: $1/4$
The probability that both dice show odd numbers is: $1/4$
The probability that one die shows even and the other shows odd is: $1/2$
This answer is for the original version of the question before the author changed the question.
An example: Let's just look at nondecreasing sequences. To deal with nonincreasing sequences, the approach is the same (and to deal with them together, you need to subtract the number of constant sequences).
You have the set $\{1,2,3\}$. Let $x_1$ be the number of $1$'s, $x_2$ be the number of $2$'s and $x_3$ be the number of $3$'s. Once the counts of $1$, $2$, and $3$ are known, the increasing and decreasing sequences are known.
Since the sequences must be of length $3$, you're considering the problem of finding $x_1\geq 0$, $x_2\geq 0$, and $x_3\geq 0$ such that $x_1+x_2+x_3=3$. This is a standard formulation of the stars-and-bars problem. The number of such combinations is
$$
\binom{3+3-1}{3}=\binom{5}{3}=10.
$$
We can check this with
\begin{align*}
(1,1,1) && (1,1,2) && (1,1,3)\\
(1,2,2) && (1,2,3) && (1,3,3)\\
(2,2,2) && (2,2,3) && (2,3,3)\\
(3,3,3).
\end{align*}
In this case, the total is $2\binom{5}{3}-3=17$ ways to be nonincreasing or nondecreasing. (10 sequences that are nonincreasing, 10 sequences that are nondecreasing, and 3 sequences that are both nonincreasing and nondecreasing - the constant ones).
In general, the formula should be
$$
\binom{2N-1}{N}-N.
$$
Best Answer
There is only one way to arrange the even numbers in increasing order. Similarly there is only one way to arrange the odd numbers in decreasing order.
Choose which four of the nine positions in our final number will be occupied by the even numbers. The remaining positions will be occupied by the odd numbers. Further, within those positions the corresponding numbers must appear in the predescribed order.
We get then a final count of $\binom{9}{4}$ different valid numbers where all even digits are in ascending order and all odd digits are in descending order. This is out of the $9!$ different equally likely arrangements of the digits that we are choosing from.
Thus, taking the ratio, our probability is:
$$\dfrac{\binom{9}{4}}{9!} \approx 0.0003472\dots$$