Here's 31 weighings. I'll only use one-vs-one weighings.
One can sort 5 elements in 7 comparisons. Split the people into 3 groups of 5 and sort each one (21 weighings). Sort the three medians of the three groups by comparing each pair (3 weighings). Call these groups the light, medium, and heavy group by their sorting.
Now, note that the median of the bottom group has eight people has eight people heavier them them: the two heavier people in their group, and the three median-and-above people in each pf the other two groups. The means they are below median, as are the two below-median people in the light group. Likewise, we know three above-median people.
We can therefore eliminate these six people from considering, and find the median of the nine remaining people, which are sorted into a group of five and two-groups of two.
Note that we can merge two sorted groups of size $a$ and $b$ using $a+b-1$ weighings by comparing the two heaviest people in each group, removing the heavier one as the maximum, and repeating. So, one can merge the two sorted lists of two into a sorted list of three (3 weighings).
At this point, we've used 27 weighings, and could spend 8 weighing to merge the two groups of 4 and 5 and have the nine sorted, for a total of 35.
But, that's overkill because we just need the median of the nine. Compare the median of the group of 5 and the median of the group of 4, rounding down for the four (1 weighing). Whichever way the result is, the two heavier people of the median-heavier group are above median, and likewise for the lighter people. Eliminating these, we're left with two sorted people and three sorted people.
Again, we could merge-sort in 4 weighings, but we instead do it in 3. Compare the median of the three to each of the two (2 weighings). If the results are opposite, the median of the three is the median, and we're done. Otherwise, say WLOG both are lighter. Compare the heavier of the two to the lightest of the three (1weighing). If he's lighter, the lighter of three is the median. Otherwise, he's the median, as we know two people lighter than him and two people heavier.
That's a total of 31 = 21 + 3 + 3 + 4 weighings.
[A computer search by Kenneth Oksanen found that the optimal number of comparisons (one-vs-one weighings) is between 24 and 28 inclusive, though an exact answer is not known.]