Each interaction decreases each number of party members by 1 $\mod 4$, and since the ending position is $0,0,0,1\mod4$ we must have one of the numbers be one more than each of the other three. In our case, it's $3,3,3,0$ so it should be possible. In fact, three conversions to party D, two to party C, and twenty six to party B give party B everyone.

And part (b)

For part a): The answer is yes and they will belong to party $B$. Define a quadruple $(a,b,c,d)$ where $a,b,c,d$ are the number of members of parties $A,B,C,D$ respectively. Construction for party $B$. $(31,28,23,19) \rightarrow (30,27,26,18) \rightarrow (29,26,29,17) \rightarrow (28,29,28,16) \rightarrow (27,28,27,19) \rightarrow (26,27,26,22) \rightarrow (25,26,25,25)$ and from this point repeating 25 times we get $(0,101,0,0)$. Now for the reason why only party $B$. Notice that at each move the number of members of all parties decreases by $1$ modulo $4$. In the starting configuration we have $(3,0,3,3)$ modulo $4$ and at ending we will only get $(0,1,0,0)$ modulo $4$. So, only $B$ is the desired party.

Let us consider the sum $b_i+2c_i+3d_i \pmod 4$, where $b_i$ is the number of people in party $B$ at some point, and so on. By checking every possible combination, this sum changes by precisely $2 \bmod 4$ every move. Furthermore, this quantity is initially odd, which means that only $B$ and $D$ are possible candidates. Similarly, considering the sum $b_i+2d_i+3c_i$ rules out $D$ as well, so only $B$ is a possible candidate. To see that $B$ indeed works, perform the operation $ABC \to D$ three times, $ABD \to C$ twice, and then $ACD \to B$ 26 times.

Solved with SatisfiedMagma. Solution. The answer is only party $B$. Let an ordered tuple $(A,B,C,D)$ denote the number of people at any point of time. The following sequence justifies that $B$ works. \[(31,28,23,19) \mapsto (30,27,26,18) \mapsto (29,26,29,17) \mapsto (28,29,28,16) \mapsto (27,28,27,19) \mapsto (27,28,27,19) \mapsto (26,27,26,22) \mapsto (25,26,25,25)\]Just keep removing members from $A,C,D$ and we will eventually reach $(0,101,0,0)$ as desired. For the other direction, look at the tuple modulo $2$. Initially the tuple $(1,0,1,1)$. It is easy to see that the tuple will change the parity of each element in every iteration. So, the tuple can only reach $(1,0,1,1)$ or $(0,1,0,0)$ in modulo $2$. This clearly means that $101$ can only appear in the $B$ spot and not anywhere else. $\blacksquare$

I will use a strategy for a problem that previously solves: If there are the same number of members in three parties, then at some point all the kidnapped people will belong to the 4th party. Note that B can never have the same number of members as 2 other parties. Only parties A=C=D can be equal. Party B can never be equal to 3 parties, this is due to: 1) a=If exactly three people from party B C D get together 2) b=If exactly three people from party A C D get together 3) c=If exactly three people from party A B D get together 4) d=If exactly three people from party A B C get together Then the cardinality of each one depends on the following equation: 1.1) A= 31 +3a -b-c-d 1.1) B= 28 +3b -a-c-d 1.1) C= 23 +3c -a-b-d 1.1) D= 19 +3d -a-b-c Suppose A=B then \[31 +3a -b-c-d = 28 +3b -a-c-d \]\[3 +4a -4b = 0\]And the above is not possible due to parity, this is done symmetrically with B = C, B = D and we will arrive at the same contradiction. Therefore, the only way to stay the same party the kidnapped people is for A=C=D to occur at some point. \[2+a-c=0 (A=C)\]\[3+a-d=0 (A=D)\]\[1+c-d=0 (C=D)\]A solution solution is (a,c,d) =(0,2,3). So a solution is with (a,b,c,d) = (0,26,2,3) $(31,28,23,19) \rightarrow (30,27,26,18) \rightarrow (29,26,29,17) \rightarrow (28,25,28,20) \rightarrow (27,24,27,23) \rightarrow (26,23,26,26) \rightarrow (0,101,0,0) $ Therefore, all the kidnapped people can only be in party B. \(\blacksquare\)

Umm ig I do not see the following invariant above. Here is a sketch for my solution for part (a). Note that $(|A| + |C| - |B| - |D|)$ and $(|A|+|D|-|B|-|C|)$ are both invariant under $\mod 4$. Thus checking the initial values of each of these invariant, we get that only $B$ can work. Now for the construction, let $A$ denote a move on party $A$. The following set of moves does it, $D\rightarrow D\rightarrow D\rightarrow C\rightarrow C\rightarrow B\rightarrow B\rightarrow \cdots \rightarrow B$ works.

\begin{claim} We claim all the kidnapped people can end up at party B. \end{claim} Let the notation $(a,b,c,d)$ denote the number of peple in party $A$,$B$,$C$,$D$ respectively. The construction for group B is as follows: \begin{align} & \text{Convert to party D : 4 time} \\ & \text{Convert to party C twice} \\ & \text{Convert to party B : 26 times} \\ \end{align}For proof that this is the only such possible endpoint for the tuple, take each element of the tuple modulo $4$. Particularly, each activity decreases the number of people in each party by $1 \pmod 4$. By performing the above construction on the initial tuple modulo $4$ , $(3,0,3,3)$ we get $(0,1,0,0)$ which obviously works. $\square$

I do,not know about the people but the wizard would go to jail

We claim only party $B$ can have all the people at any point in time. Suppose $a$ conversations occur converting $3$ people to party $A$. Define $b$, $c$, $d$ similarly. Then party $A$ will have $$31 + 3a - b - c - d = 31 + 4a - (a + b + c + d)$$people remaining, and similar results hold for other parties. This implies that if $3$ parties have the same number of people, their original number of people in $\pmod{4}$ should be equal. This only occurs with parties $A$, $C$, $D$ so $B$ is the only party that can have all the people. To show sufficiency note that $(a, b, c, d) = (0, 26, 2, 3)$ satisfies the equations hence it is possible.

All of the people can belong to party $B$ only. Note that $\pmod 4$, the parties will have to have 1 0 0 0 people at the end (some permutation of that). However, the $\pmod 4$ values never change in respect to each other (after each move, the $\pmod 4$ values all go down by $1\pmod 4$). The original values are 0 3 3 3 people $\pmod 4$. The party that is $0\pmod 4$ (one ahead of everyone else) will be the party that everyone belongs to. The construction is $$ABC\to D\times 3, ABD\to C\times 2, ACD\to B\times 26.$$