We write such a five-digit positive integer with digits $V W X Y Z$. We want to count the number of ways of assigning $1, 3, 5, 7, 9$ to the digits $V, W, X, Y, Z$ in such a way that the given properties are obeyed. From the given conditions, $W > X, W > V, Y > X$, and $Y > Z$. The digits 1 and 3 cannot be placed as $W$ or $Y$, since $W$ and $Y$ are larger than both of their neighbouring digits, while 1 is smaller than all of the other digits and 3 is only larger than one of the other possible digits. The digit 9 cannot be placed as $V, X$ or $Z$ since it is the largest possible digit and so cannot be smaller than $W$ or $Y$. Thus, 9 is placed as $W$ or as $Y$. Therefore, the digits $W$ and $Y$ are 9 and either 5 or 7. Suppose that $W = 9$ and $Y = 5$. The number is thus $V 9 X 5 Z$. Neither $X$ or $Z$ can equal 7 since $7 > 5$, so $V = 7$. $X$ and $Z$ are then 1 and 3 or 3 and 1. There are 2 possible integers in this case. Similarly, if $Y = 9$ and $W = 5$, there are 2 possible integers. Suppose that $W = 9$ and $Y = 7$. The number is thus $V 9 X 7 Z$. The digits $1, 3, 5$ can be placed in any of the remaining spots. There are 3 choices for the digit $V$. For each of these choices, there are 2 choices for $X$ and then 1 choice for $Z$. There are thus $3 \times 2 \times 1 = 6$ possible integers in this case. Similarly, if $Y = 9$ and $W = 7$, there are 6 possible integers. Overall, there are thus $2 + 2 + 6 + 6 = 16$ possible integers.