The comments ask for a general approach to weighted medians. In the approach that makes sense to me, the weights end up the same as for weighted means.
The following result on means is a straightforward constrained optimization (e.g. here):
Suppose we have $n$ different methods of measuring the same quantity, and the sample mean $M_i$ from method $i$ has mean $\mu$ and variance $V_i$. Then the minimum-variance weighted average of those sample means is $$\frac{\sum M_i\, / \, V_i}{\sum 1 \, / \,V_i}$$
The sample medians, as asked about and pointed out in the question, have normal distributions. So a weighted average of sample medians will also be distributed normally. The variances will be proportional to the squares of the interquartile ranges, and minimizing the overall interquartile range will have the same result as minimizing the overall variance. This leads to the following result, parallel to the statement on means but now entirely in quantile terms:
Suppose we have $n$ different methods of measuring the same quantity, and the sample median $M_i$ from method $i$ has median $\mu$ and interquartile range $r_i$. Then the minimum-interquartile-range weighted average of those sample medians is $$\frac{\sum M_i\, / \, r_i^2}{\sum 1\, / \, r_i^2}$$
If the goal is instead to take a weighted median of the samples (or of their sample median distributions), then we are looking for the mixture $R$ of the distributions $X_i$ (or of their sample median distributions) so that the sample median of $R$ has minimal variance.
Since the variance of the sample median of $R$ is inversely proportional to the pdf of $R$ at the median, we are looking for the mixture $R$ with highest possible pdf at that median. Assuming that all the $X_i$ have the same median (as in the post, and as in the case where they are all normal or all symmetric about the origin), the optimal mixture of $X_i$’s will be exclusively composed of the one $X_i$ which has the highest pdf at its median.