Suppose we keep on removing weights from the scale on the tilted side, and instead of stopping when the balance gets in the equilibrium position with some weights still left on the scales, remove another weight from the left-side scale and keep on removing weights from the tilted side again until there are no more weights to be removed. Then the total number of ways of removing weights is given by $(4!)^2 = 576$, which is equal to the total number of ordering the removals of weights from each of the scales. The desired number we seek as the answer to the problem is obtained by subtracting from $576$ the number of ways where equilibrium is reached with some weights still remaining on the scales.
Suppose we denote by $r$ the ratio (taken as no less than $1$) of the numbers of weights left on the two scales when equilibrium is reached, then $r$ is no more than the ratio $\frac{29}{22}$ of the heaviest weight to the lightest, since the product of the number of weights and the average weight of the weights must be equal for the both scales, when equilibrium is attained. Since the numbers of the weights left are no more than $4$, the smallest possible value for $r$ is, if not equal to $1$, $\frac{4}{3}$. Since $\frac{4}{3} > \frac{29}{22}$, we conclude that $r = 1$, which means that if equilibrium is achieved, then there must be a same number of weights left on both scales.
Since the value of each weight on the left scale is even, the total weight on each of the scales when equilibrium is achieved must be even. Since the value of each weight on the right scale is odd, the number of weights on each scale must be a same even number; in fact, we can see that both scales must have $2$ weights when equilibrium occurs. We can check easily that the following $6$ cases are the only possibilities:
$$\begin{aligned} 22 + 26 &= 23 + 25, & 22 + 28 &= 23 + 27, & 24 + 26 &= 23 + 27, \\ 24 + 28 &= 23 + 29, & 24 + 28 &= 25 + 27, & 26 + 28 &= 25 + 29. \end{aligned}$$
For each of these $6$ possibilities there are $2$ occasions before equilibrium is achieved and $2$ occasions after it is achieved, when you can make a choice of deciding which of the two remaining weights to remove from the scales, and therefore, the total number of ways of removing weights to attain equilibrium with some weights still remaining on the scales is $6 \times 2^4 = 96$ and the desired answer to the problem is $576 - 96 = 480$.