Let $n=2m+1$. Each candidate appears in the vote matrix $2m+1$ times. (a) For $k=3$, when $\overline{l} = 2$, let candidate $a$ be the first and $b$ be the second. By definition, $l_a = 2$. There are $4m+2$ votes in the first two rows. $a$ has at most $2m+1$ votes into these two rows. So at most $4m+2 - (2m+1) = 2m+1$ votes are for other two candidates. By pigeonhole principle, in the first two rows, $b$ has at least $\left \lceil \frac {2m+1}2 \right \rceil = m+1$. Then this means $l_b=2$. In that case, there is no $\overline{l}$. So for $k=3$, it should be that $\overline{l} < 2$. For $\overline{l} = 1$, let $a$ be the first, $b$ be the second. \[\begin{array}{ccccccccc} &1&2&\dots&m&m+1&m+2&\dots&2m+1 \\ w_1&a&a&a&a&a&b&b&b \\ w_2&b&b&b&b&b&&& \\ w_3&&&&&&a&a&a \end{array}\]The minimum total weight for candidate $a$ is $(m+1)w_1 + mw_3$. The maximum total weight for candidate $b$ is $mw_1 + (m+1)w_2$. To represent majoritarian compromise, it should be \[(m+1)w_1 + mw_3 > mw_1 + (m+1)w_2 \Rightarrow w_1-w_2 > m(w_2-w_3).\]When we choose $w_1>w_2=w_3$, the above equality holds for every $m>0$. So it is possible to find a system of weights to represent majoritarian compromise. (b) For $k>3$, when $\overline{l} = 1$, \[\begin{array}{ccccccccc} &1&2&\dots&m&m+1&m+2&\dots&2m+1 \\ w_1&a&a&a&a&a&b&b&b \\ w_2&b&b&b&b&b&&& \\ \dots \\ w_k&&&&&&a&a&a \end{array}\]\[(m+1)w_1 + mw_k > mw_1 + (m+1)w_2 \Rightarrow w_1-w_2 > m(w_2-w_k).\] For $k=4$, when $\overline{l} = 2$, At most $3m$ of $4m+2$ votes in the first two rows can belong to candidates other than the first. In that case, the first candidate $a$ has at least $m+2$ votes in the first two rows. \[\begin{array}{ccccccccc} &1&2&\dots&m&m+1&m+2&\dots&2m+1 \\ w_1&b&b&b&b&c&d&d&d \\ w_2&a&a&a&a&a&a&c&c \\ w_3&&&&&b&b&b&b \\ w_4&&&&&&&a&a \end{array}\]If we compare minimum weights for $a$ and maximum weight for $b$, we get \[(m+2)w_2 + (m-1)w_4 > mw_1 + (m+1)w_2 \Rightarrow w_2 - w_4 > m (w_1 - w_4).\]For $k=4$ and $\overline{l} = 1$, we had $w_1 - w_2 > m(w_2-w_4)$. If we merge these two inequalities, we will get \[\frac{w_1-w_2}{m} > w_2-w_4 > m(w_1 - w_4) \Rightarrow 1 \geq \frac{w_1-w_2}{w_1-w_4}>m^2.\]This concludes for $m>1$, such a system of weights does not exist for $k=4$. For $k>4$, when $\overline{l} = 2$, \[\begin{array}{ccccccccc} &1&2&\dots&m&m+1&m+2&\dots&2m+1 \\ w_1&b&b&b&b&c&d&d&d \\ w_2&a&a&a&a&a&&c&c \\ w_3&&&&&b&b&b&b \\ w_k&&&&&&a&a&a \end{array}\]\[(m+1)w_2 + mw_k > mw_1 + (m+1)w_3 \Rightarrow m(w_2-w_3)>m(w_1-w_k)-(w_2-w_3)\]If we merge above inequality with the inequality for $\overline{l}=1$, we will get \[w_1-w_2 > m(w_2-w_k)\geq m(w_2-w_3)>m(w_1-w_k)-(w_2-w_3)\\ \Rightarrow w_1 -w_2 + w_2 - w_3 > m(w_1-w_k) \Rightarrow w_1-w_3 > m(w_1-w_k) \\ \Rightarrow 1 \geq \frac{w_1-w_3}{w_1-w_k}>m.\]This concludes for $m>1$, such a system of weights does not exist for $k>4$.

A very similar solution: $a)$ Let the candidates be $a,b,c$. First, we claim that $\bar{l}=1$. To see this, we begin by proving $\bar{l}\leq 2$. Indeed, note that, $a_1+b_1+c_1=n$ and $a_2+b_2+c_2=n$, hence, $(a_1+a_2)+(b_1+b_2)+(c_1+c_2)=2n$. Therefore, at least one of $a_1+a_2$, $b_1+b_2$, or $c_1+c_2$ exceeds $2n/3>n/2$. Hence, there is an $i\in\{a,b,c\}$ such that, $\ell_i=2$, for any voter-candidate profile. Now, suppose that, $\bar{l}=2$. In this case, assume without loss of generality that $a$ is the unique winner. We have, $a_1+a_2>n/2$, $b_1+b_2<n/2$, and $c_1+c_2<n/2$. Furthermore, $a_1+b_1+c_1+a_2+b_2+c_2=2n$. With this, we obtain that, $a_1+a_2>n$, which is a contradiction (since, no voter puts $a$ in his list at both $1^{st}$ and $2^{nd}$ location, $a_1+a_2$ amounts for different people, which is clearly less than $n$). Now, $\bar{l}=1$. We clearly have, $a_1>n/2$, and $b_1,c_1<n/2$. Furthermore, we also have $\sum_{i=1}^3 a_i=\sum_{i=1}^3 b_i=\sum_{i=1}^3 c_i=n$. Hence, it suffices to take $w_1>1$, $w_2=w_3=1$. Note that, in this case, $w_a=(w_1-1)a_1+a_1+a_2+a_3=(w_1-1)a_1+n$. Similarly, $w_b=(w_1-1)b_1+n$, and $w_c=(w_1-1)c_1+n$, and clearly, since $a_1>b_1,c_1$, $w_a>w_b,w_c$. As a side note, observe that, since $n$ is odd, all the strict inequalities above are valid, hence the argument. $b)$ The key idea is that, we will try to arrive at a contradiction, involving the weights, $w_1,\dots,w_k$. In what follows, always assume $a$ is the winner, and let $n=2m+1$. First, if $\bar{l}=1$, then, $a_1\geq m+1$. Suppose, $b$ is a second candidate. We have, $b_1\leq m$. Now, the minimum weight, that $a$ can has, is at least $\sum_{i=1}^k w_ia_i\geq (m+1)w_1+mw_k$ (consider a voting profile, where exactly $m+1$ voters vote for $a$ as their first choice, and the remaining $m$ put him as their last choice. This clearly is an admissible matrix, and furthermore, $a$ is the winner, based on majoritarian compromise. In this scenario, according to the statement, $a$ must have the largest weight). Now, for a second candidate, the best bet is as follows. $b_1=m$, and $b_2=m+1$ (the remaining people). This gives, $ w_b=mw_1+(m+1)w_2$. Now, $$ w_a>w_b\implies w_1-w_2>m(w_2-w_k). $$Now, we turn our attention to a voting profile, where $\bar{l}=2$ (such profiles clearly exist). Assume that $k=4$. We have $a_1\leq m$, and $a_1+a_2\geq m+1$. Furthermore, $\sum_{i=1}^2 (a_i+b_i+c_i+d_i)=2n=4m+2$. If, $a_1+a_2=m+1$, then, we have, $(b_1+b_2)+(c_1+c_2)+(d_1+d_2)=3m+1$, thus by Pigeonhole, at least one of $\xi = b,c,d$ satisfy $\xi_1+\xi_2\geq m+1$, contradicting with the fact that $|\{a\in A:l_a=\bar{l}\}|=1$. Hence, $a_1+a_2\geq m+2$. With this, we now consider the smallest possible weight for such a candidate-vote profile. Assume exactly $m+2$ people put $a$ as their second choice, and the remaining $m-1$ put him as their last choice. The weight becomes $(m+2)w_1+(m-1)w_4$. Now, for the second candidate, at most $m$ people put him as their first choice. Assume exactly $m$ people put him. His weight, in this case, is strictly less than $mw_1+(m+1)w_2$. Thus, we have, $(m+2)w_2+(m-1)w_4>mw_1+(m+1)w_2\implies w_2-w_4>m(w_1-w_4)$. Hence, $$ w_2-w_4>m(w_1-w_4)>m(w_1-w_2)>m^2(w_2-w_k)>m^2(w_2-w_4), $$gives a contradiction, when $m>1$. Finally, suppose that, $k>4$. With the same logic as above, $a_1+a_2\geq m+1$. Consider a profile, where exactly $m+1$ people put $a$ as their second choice, and the remaining $m$ people put him as their last choice. Hence, the weight of $a$ becomes $(m+1)w_2+mw_k$. Now, consider another person, where exactly $m$ people put him as their first choice, and $m+1$ people as their third choice. We have, his score $mw_1+(m+1)w_3$. We have, $$ (m+1)w_2+mw_k>mw_1+(m+1)w_3. $$However, combining this, with $w_1-w_2>mw_2-mw_k\iff w_1>(m+1)w_2-mw_k$, we have, $$ w_1>mw_1-2mw_k+(m+1)w_3\implies 2mw_k>mw_1+(m+1)w_3, $$which is a clear contradiction.

I happened to learn that majoritarian compromise is an electoral system introduced by the late Turkish economist, Prof. Murat Sertel. See here for a relevant paper. This problem has most likely been proposed by Prof. Semih Koray, who also works in similar areas with Dr. Sertel.