Specially for Pascual By induction: if we have $2n$ men, then thre are at most $n$ voitings. Suppose the contrary. Let $a_1$, $a_2$, ..., $a_{n}$ be number of people excluded after 1st, 2nd, ..., 15th voiting respectively. Let $b_i$ be number of competences from non-excluded people after 1st, ..., (i-1)th voiting, which posses man excluded in $i$-th voiting. If $a_1\geq 2$ then, obviously, we are done. So we will assume $a_1=1$. Consider the man $A$, which was excluded in $i$-th voiting. Since he wasn't excluded in $i-1$th voiting we conclude \[b_i+a_{i-1}\geq \frac{2n-a_1-a_2-...-a_{i-2}}{2},\] but as he was excluded in $i$th voiting we can write \[b_i<\frac{2n-a_1-a_2-...-a_{i-1}}{2}.\] It follows \[\frac{2n-a_1-a_2-...-a_{i-1}-1}{2}\geq b_i \geq \frac{2n-a_1-...-a_{i-2}-2a_{i-1}}{2}.\qquad \eqno (1)\] It is clear there are $i$ s.t. $n\geq i-1\geq 2$ and $a_{i-1}=1$ (otherwise after $n$ votings there is at most 1 man in jury, and he cannot be excluded). Let $m>2$ be minimal such $i$. From $(1)$ we conclude $b_{m}=\frac{2n-a_1-a_2-...-a_{m-1}-1}{2}$, so $a_1+a_2+...+a_{m-2}$ is even. Due to choice of $m$ it means $a_1+...+a_{m-2}\geq 2(m-2)$. And, obviously, we are done.