One can find the lightest ball $13$ via direct elimination by pairs: we make $8$ pairs and determine the $8$ lightest on each pair, now we make $4$ pairs and determine the lightest on each pair, then we select the two lightest and finally, the ball with the minimum weight. This takes $8+4+2+1=15$ attempts. Ball $14$ is among the ones eliminated by ball $13$ in the process. There are $4$ of them; $14$ is the lightest one, and it can be found by direct elimination again with $2+1=3$ attempts.

Observe now that $28$ is the only ball heavier than $13$ and $14$ combined. Likewise $27$ is the only ball with the same weight as $13$ and $14$ combined. In addition, ball $28$ is among the eight losers in the first eight uses of the balance. Let them be $B_1, \ldots, B_8$. (It could be less if $14$ was eliminated by $13$ in the first eight uses of the balance.)

For each $i = 1, \ldots, 8$ compare ball $B_i$ with the group $13$, $14$. One of these eight attempts will find ball $28$. If there is equilibrium at one of the seven remaining attempts, ball $27$ is found too. Otherwise $27$ is a winner in the first round, which is possible only if it was compared with $28$ at the first round. Therefore, because $28$ is already known, so is also $27$; no further attempts are needed. The task is solved with $15+3+8=26$ attempts.