If #whites>#blacks you can find consecutive 3 white and 1 black vertex. Between black blocks in the configuration: 1) If at least 3 consecutive white vertices exist, then you are done. 2) If white blocks can have at most 2 vertices: obviously, there exists a white block with 2 elements and a black block with 1 element followed by at least one white vertex. (Otherwise, #blacks becomes greater than #whites) Afterward, you can delete 3 white 1 black until 32 white 32 black and you need 15 additional squares. Do induction for 2n white 2n black for finding n-1 quadrilaterals. Base cases are trivial(1,2). If you can delete 3 white and 1 black vertex then you can delete 3 black and 1 white as well since #white becomes less than #black. As in the aforementioned claims, we get 2 cases: white and black vertices alternating with 2 vertices in each block or white and black vertices alternating with 1 vertex in each block Construction for them is as above.

Note that all that matters is the order of the vertices. It suffices to show that we can choose quads s.t. we are left with $\leq 4$ squares not in a weird quad. Lemma 1: In an $n$-gon, if there are $W$ whties and $B$ blacks, then if $W>B>0$, then there exists 4 adjacent vertices with 3 whites and 1 black. Proof: Firstly, note that there must be some adjacent 4 with more white than black b/c otherwise cyclically summing would give $W\leq B$, contradiction. AFTOSC there are no 3 white, 1black sets of 4, thus there must exist a set of 4 with all whites. Since $B>0$, there exists a black vertex, so consider the set of 4-tuples between the WWWW and a 4-tuple containing the black. Since the number of blacks changes by $\leq 1$ per shift, there must exist some 4-tuple with exactly 1 black, so we are done. $\square$. Lemma 2: In the 100-gon with $B=41,W=59$, there exist $\geq 9$ sets of adjacent weird quads Proof: Repeatedly applying Lemma 1 says that we will be able to find at least \[\min(41,\frac{59-41}{2})=9\text{sets of adjacents}\] Now, remove those $\geq 9$ disjoint sets of 4 adjacent vertices with 3 whites and 1 black, this contributes 9 and we are left with an $2k$-gon with $k$ blacks and $k$ whites. As a post-condition we have that each set of 4 adj. have 2 whites and 2 blacks We now do casework. Case 1: $\exists$ a pair of vertices with the same color. The $2k$-gon is forced to look like \[\cdots 2211221122(1)122112211\cdots \]it is forced because if 2 adj are both color 1, then the pair adj to that pair must be color 2, etc. Here, we select $P=(1)$, an arbitrary point $P$, and while there are $\geq 4$ unincluded points, put together the set of the two closest to $P$ clockwise, and the two closest counterclockwise, which due to parity will all be weird quads, so this case is done. \includegraphics[scale=0.3]{case1.png} Case 2: No pairs of adjacents are the same color. Then, there must exist two vertices of the same color 2 spots apart or else we can't fit all the vertices in $\frac{2k}{3}+\frac{2k}{3}<2k$. Thus, two must be 2 apart, from which everything else comes so the diagram will look like \[\cdots 1212121212(1)21212121\cdots \]\includegraphics[scale=0.3]{case2.png} Then, as long as there are $\geq 4$ unassigned vertices, place together the vertex closest to P co. clockwise, and the 3 closest from $P$ in the clockwise direction, which once again by parity reasons guarantees weird quadrilaterals, so we can assign all except 4 vertices. $\square$. Thus, we are no finished b/c no matter what we can limit the unassigneds to $\leq 4$, so there are $\geq 96$ assigned, and we will have assigned at least 24 quads with this alg, so we are done. $\blacksquare$.

We call a colour dominiant if there is more vertices of that colour than the other. We call a quadrilateral good if it is formed by four consecutive vertices on the circle and satisfies $(ii)$. We call a colouring excellent if there exists four consecutive vertices with exactly three vertices coloured by the domininat colour. For example, if White is dominant then $$W\hspace{3pt}B\hspace{3pt}W\hspace{3pt}W$$will be such a configuration. \bfblue{Claim. } There exists a good quadrilateral for each colouring with more than four vertices, and the number of black white vertices are not the same, and there exists vertices of both colours. Proof. Suppose there are $W$ white and $B$ black vertices then $W\neq B$. WLOG assume $W>B>0$. Let the vertices be $V_1,V_2,...,V_k$ in clockwise order. Let $w_i$ be the number of white vertices in $V_i,V_{i+1},V_{i+2},V_{i+3}$. NOtice that there exists some $i$ with $w_i<4$, or otherwise $B=0$, and there exists some $j$ with $w_j<2$, otherwise $W\leq B$. \newline\newline Meanwhile $|w_i-w_j|\leq 1$. So by discrete continuity there exists some $k$ such that $w_i=3$. Just pick the quadrilateral $V_iV_{i+1}V_{i+2}V_{i+3}$ and this is a good quadrilateral. $\blacksquare$ Now in each step we use the claim to select the quadrilateral, then we erase the points on the quadrilateral and continue. We are good as long as the colouring is excellent. In particular, no problem will occur if the number of Black and White vertices is different, it is easy to see that the number $(W,B)$ will change from $(59,41)$ to $(56,40)$,... and finally to $(32,32)$. Consider the first case when the configuration is not excellent. Then every $4$ consecutive vertices must consist of $2$ vertices of the same colour. In other words the colour of vertex $V_i$ and $V_{i+1}$ must be the same. Therefore, it must be of the following form: $$\begin{pmatrix}W&B&W&B&W&B&...\end{pmatrix}\text{ or }\begin{pmatrix}W&W&B&B&W&W&B&B&...\end{pmatrix}$$In either case we can form a quadrilateral satisfying (ii) by taking four of the first five points, then we erase the five points and proceed using the claim. Since there are now an odd number of vertices we can ensure the colouting is always excellent and we are done.

Unless I'm missing something, I don't think anybody has posted this one (which bypasses cases) yet: As in other solutions, if there is more B than W, we can find a consecutive 3B1W, and similar for more W than B. Now, delete one of the white vertices so there are 41 black vertices and 58 white vertices. Do the following repeatedly: - If there are more white than black remaining, remove a 3W1B - If there are more black than white remaining, remove a 3B1W Note that there can't be equal amounts of each color for parity reasons, so one of these is always possible. It's simple to see there are below $4$ remaining vertices when this terminates, so we are done. @below if you're referring to this solution, the idea was removing one of the vertices before starting the process, so there's an odd total number of vertices, so you wouldn't get to 32 32

why there can't be equal amounts of same colour?If I am not mistaken you will get 32 w 32 b

To start, delete one white vertex. We will repeatedly delete quadrilaterals whose 4 vertices are consecutive on the boundary. Key Claim: Assume at least $1$ vertex of each color remains. Of those vertices that remain, we can always find $4$ consecutive vertices on the boundary of which $3$ are of the majority color remaining, and $1$ is of the minority color remaining. Proof: First note that the majority and minority colors are well-defined since we deleted one white vertex, which means $99$ vertices remain, and every time we delete a quadrilateral, there will still be an odd number of total vertices remaining. WLOG assume that white is majority color. First we claim there must exist 4 consecutive that are either 4 white or 3 white 1 black. If no such substring exists, then every substring of length 4 has at least as many whites as blacks. Let $x_i$ is 1 if $i$ white and $-1$ if $i$ black (and indices cycle). We know $x_i+x_{i+1}+x_{i+2}+x_{i+3} \le 0$ for all $i$. Summing over all $i$, \[ 0\ge \sum_{i} (x_i+x_{i+1}+x_{i+2}+x_{i+3}) = 4\sum_i x_i. \]However, $\sum x_i>0$ since there are more whites than blacks (can't even be equal since odd total vertices), contradiction. Now, if we have 4 consecutive whites, simply shift by 1 repeatedly, and we will eventually get $3$ whites and $1$ black. This holds since we assumed there is at least $1$ of each color. $\blacksquare$ Delete the quadrilateral generated by the claim. We claim that we can repeat this process at least $24$ times total. The distribution of (whites, blacks) is: \begin{align*} (58,41) &\to (55, 40) \to (52, 39) \to (49, 38) \to (46, 37) \to (43,36) \\ &\to (40,35)\to (37,34)\to (34,33) \to (31, 32) \to (30,29) \\ &\to (27,28) \to (26,25)\to (23,24) \to (22, 21) \to (19,20)\\ &\to (18,17) \to (15,16) \to (14,13) \to (11,12) \to (10,9) \\ &\to (7,8) \to (6,5) \to (3,4) \to (2,1). \end{align*}This has exactly 24 deletions, and we are done.

I really liked the idea of making the vertices odd in order to always have imbalance in number of white and black vertices. But below is a proof in that original world of $4k$ vertices only (i.e. even vertices ; since I was unable to find that ingenious idea of making the vertices odd ): We will a call a vertex active if it is not part of a quadrilateral yet and we will call quadrilateral good if any segment joining two active points does not intersect the quadrilateral. Claim: If there are more active vertex of color black than of white, then there is a good quadrilateral with 3 black and 1 white vertex (and vise versa). proof: Not that hard (this claim is also proven in some previous posts, so detailed proof could be referred from there ; we are omitting the full proof here since this is not a key claim in our proof). $\square$ First we put as many good (and disjoint) quadrilaterals as we can in our collection of quadrilaterals in such a way that if we have more active black vertices than white, then we give preference to the good quadrilaterals with 3 black and 1 white vertex instead of a quadrilateral with 1 black and 3 white vertex (and vise versa). This is done in order to avoid shortage of vertex of some particular (like, we don't want to reach some situation (for example) where there are only $11$ black vertices and $49$ white vertices). After that, we have a situation where there are no more good quadrilaterals. Let $P_1,P_2,\dots,P_{4n}$ be the active points (in that order) at that time. Claim: Among $P_1,P_2,\dots,P_{4n}$, either the black and white will alternate or there will be two black-two white-two black- two white and so on. proof: Assume the contrary. Then there must be four points $P_i,P_{i+1},P_{i+2},P_{i+3}$ (indices modulo $4n$) such that $P_i,P_{i+1}$ have the same color and $P_{i+2},P_{i+3}$ have different colors. Then it is easy to see that quadrilateral $P_iP_{i+1}P_{i+2}P_{i+3}$ is also good, which is a contradiction. This proves our claim. $\square$ Also, we just have to prove that we can choose $n-1$ pairwise disjoint quadrilaterals with vertices among $P_1,P_2,\dots,P_{4n}$ with the second condition. Now there are just two type of cases to handle: CASE 1: $P_1,P_3,\dots,P_{4n-1}$ were colored black and $P_2,P_4,\dots,P_{4n}$ were white. Note that following is such a set of $n-1$ quadrilaterals (the intuitive reason being that we waste the vertex $P_{4n}$ only) : $$P_1P_2P_3P_{4n-1} ~,~ P_4P_5P_6P_{4n-2} ~,~ P_7P_8P_9P_{4n-3} ~,~ P_{10}P_{11}P_{12}P_{4n-4} ~,~ \dots ~,~ P_{3n-5}P_{3n-4}P_{3n-3}P_{3n+1}$$ CASE 2: $P_1,P_2,P_5,P_6,\dots,P_{4n-3},P_{4n-2}$ were colored black and $P_3,P_4,P_7,P_8,\dots,P_{4n-1},P_{4n}$ were colored white. Note that following is such a set of $n-1$ quadrilaterals (the intuitive reason being that we waste the vertices $P_{4n},P_{4n-1}$ only) : $$ P_1P_2P_3P_{4n-2} ~,~ P_4P_5P_6P_{4n-3} ~,~ P_7P_8P_9P_{4n-4} ~,~ P_{10}P_{11}P_{12}P_{4n-5} ~,~ \dots ~,~ P_{3n-5}P_{3n-4}P_{3n-3}P_{3n} $$ This completes the proof of the problem. $\blacksquare$ Comment: I think the main difficulty in this solution was to observe the second claim and finding appropriate constructions for both cases.

To handle the disjoint condition, we present a sort of algorithmic approach where we choose a quadrilateral with $4$ adjacent vertices, erase those $4$ vertices from the $n$-gon, and repeat. Let $B$ and $W$ denote the current number of black and white vertices, respectively. Lemma $1$: If $B > W > 0$, there exist $4$ adjacent vertices, exactly $3$ of which are black. The reverse holds if $W > B > 0$. Proof: Suppose, towards contradiction, that there does not exist $4$ such vertices. Since all vertices are not black, it is trivial to see that if there are $4$ adjacent back vertices, then there are also $4$ with $3$ black and $1$ white. Hence, among any $4$ adjacent vertices, at least as many are white as black. But summing over all sets of $4$ adjacent vertices, we obtain $4W\ge 4B$, absurd, and we have a contradiction. Now discard $3$ arbitrary vertices of the $100$-gon. It is easy to see that, if we keep finding and erasing quadrilaterals as in the lemma, $B$ and $W$ will always be distinct and neither can equal $0$ until there is only $1$ vertex left, at which point we are done.

Claim: If the number of white vertices isn't equal to the number of black vertices, then there exists a set of $4$ consecutive vertices where $3$ are of one color and the remaining is of the other color. If the number of white vertices is greater than the number of black vertices we can find a quadrilateral with exactly $3$ white vertices, and the opposite also holds. Proof: If there doesn't exist a set of $4$ consecutive vertices that satisfy the conditions in our claim, then each of the sets either have $4$ black/white vertices, or $2$ vertices of each color. It is easy to see that the former condition will not hold for our entire polygon if the number of vertices of each color is greater than $1$. The other condition the number of vertices of each color are equal, so we are done. Delete an arbitrary white vertex. We are now left with $58$ white vertices and $41$ black vertices. After finding a quadrilateral that satisfies our criteria that has $4$ consecutive vertices, let us delete all of those vertices. We can delete $9$ "white" quadrilaterals until we have $31$ white vertices and $32$ black vertices. Then, we can alternate deleting "black" quadrilaterals and "white" quadrilaterals until we delete $24$ quadrilaterals. By our claim, we can always find such a quadrilateral since the number of vertices of each color are never equal. $\blacksquare$

Say that the white vertices in the given 100-gon each have a value of $1$ and the black vertices have no value. Consider all possible quadrilaterals formed by taking $4$ consecutive vertices in the 100-gon. There are 100 such quadrilaterals. Suppose that no such quadrilateral contained 3 white vertices. Then the total value of all such quadrilaterals must be at most $2 \cdot 100 = 200$, but this must be equal to $4 \cdot 59 = 236$ as each white vertex was counted four times. This contradicts that no consecutive quadrilaterals have $3$ white vertices (if there are $4$ consecutive white vertices then consider the leftmost 3 and the one to the left of that, and if those 4 are also white, then keep going, eventually you will encounter a black vertex, so there must be 3 consecutive white vertices followed by a black one as we wanted). Now draw one such quadrilateral containing $3$ white vertices and look at the rest of the 96 vertices. There are $56$ white vertices and $40$ black vertices. Continue repeating this contradiction $9$ times until we have $32$ black and white vertices each. Call a quadrilateral good if it has three vertices of the same color. It is pretty intuitive to see that it is hardest to divide a polygon into good quadrilaterals if the number of black and white vertices are equal or close to equal i.e. if one can color the vertices of an n-sided polygon black and white such that the number of black and white vertices are equal such that the polygon can be divided into $k$ good quadrilaterals, then one can divide any n-sided polygon into $k$ good quadrilaterals with any other distribution of black and white vertices. So we may assume that in the remaining $64$ vertices, there don't exist any consecutive $4$ points that form a good quadrilateral. If there did, we can just reduce the problem to a simpler one, so its sufficient to solve it when there do not exist any consecutive good quadrilaterals. Here we have 2 cases (observe that if a certain distribution of black and white vertices apart from the following 2 existed, then there must be a consecutive good quadrilateral: (i) The vertices forms a repeating $BWBWB \cdots BW$. If this happens, then label the vertices $1, 2, 3, \cdots 64$ in order. Consider the quadrilaterals $(1, 2, 3, 5)$, $(6, 62, 63, 64)$, $(61, 7, 8, 9), \cdots $ (it may be easier to see the pattern if you draw it on paper) and if you continue this for the remaining points, there will be $3$ unmatched vertices at the end, and combined with the vertex with label $4$, there are a total of $4$ vertices that are not part of a good quadrilateral, which means there are a total of $24$ disjoint good quadrilaterals drawn, so we are done. (ii) The vertices form a repeating $BBWWBBWWBBWW \cdots BBWW$ pattern. Then use the same quadrilaterals as above but ensure that $1$ and $2$ are of the same color. Again, $4$ vertices will not be part of a good quadrilateral for a total of $24$ disjoint good quadrilaterals.

Sorry for the question, but why selecting $7$ quadrilaterals with $3$ black vertices and $17$ quadrilaterals with $3$ white vertices doesn't work? In total they have $7 \times 3 + 17 = 38$ black vertices and $7 + 17 \times 3 = 58$ white vertices.

To begin, delete a white vertex. Claim: Suppose that most vertices are white and there is at least 1 vertex. Then, we can find a consecutive group of 3 white vertices and 1 black vertex. The same holds when black has the majority. Proof: If there exists a run of at least 4 consecutive white vertices, we can find a 3w1b group towards its end. Otherwise, assume that every group of 4 consecutive vertices has 2w, 1w, or 0w; a global sum gives that at most half of the vertices are white, a contradiction. Proceed with a process where after we find each quadrilateral, we ignore it and start finding another quadrilateral within the remainder of the 99-gon. We can iterate it 24 times [work needed] to arrive at our desired result.

BolaOne wrote: Sorry for the question, but why selecting $7$ quadrilaterals with $3$ black vertices and $17$ quadrilaterals with $3$ white vertices doesn't work? In total they have $7 \times 3 + 17 = 38$ black vertices and $7 + 17 \times 3 = 58$ white vertices. Well, it would work, but you have to guarantee that the quadrilaterals are disjoint (basically, they don't overlap)

Solution. Claim 1. If there are $x$ black vertices and $y$ white vertices, then if $x > y$ there are four consecutive vertices, exactly three of which are black. And vice versa. Proof. Let $B_1,\dots B_y$ be the number of black vertices between every two adjacent $y$ white vertices. If one of $B_i \ge 3$, then we are done. If $B_i + B_{i+1} \ge 3$, then we're done. Assume that $B_i + B_{i+1} \leq 2$. Then summing over $i=1,\dots y$, we have $$2(B_1+\dots +B_y) \leq 2 y$$$$2x \leq 2y$$, a contradiction. $$(58, 41) \longrightarrow (55, 40) \longrightarrow (52, 39) \longrightarrow (49, 38) \longrightarrow (46, 37) \longrightarrow (43, 36) \longrightarrow$$$$ (40, 35) \longrightarrow (37, 34) \longrightarrow (34, 33) \longrightarrow (31, 32) \longrightarrow (30, 29) \longrightarrow (27, 28) \longrightarrow $$$$(26, 25) \longrightarrow (23, 24) \longrightarrow (22, 21) \longrightarrow (19, 20) \longrightarrow (18, 17) \longrightarrow (15, 16) \longrightarrow $$$$(14, 13) \longrightarrow (11, 12) \longrightarrow (10, 9) \longrightarrow (7, 8) \longrightarrow (6, 5) \longrightarrow (3, 4) \longrightarrow$$$$ (2, 1)$$

Generalize to an $(w+b)-$gon with $w$ white vertices and $b$ black vertices. Johnny is moving around the polygon slowly and at a fixed distance from the radius, so that he can see exactly four vertices at all times. Johnny notes the number of white vertices every time he sees a new vertex. Clearly, this number is discretely continuous on the integers, which means that at some point in time Johnny must see exactly $1$ or $3$ white vertices. The only exception to this is $w=b$ or $wb=0$. Let $(w,b,c)$ denote the assertion that with $w$ white vertices and $b$ black vertices then we can find $c$ quadrilaterals. If $w>b$ then $(w-3,b-1,c-1)\implies (w,b,c)$ since you can remove four adjacent vertices with one of them being black. If $b>w$ then $(w-1,b-3,c-1)\implies (w,b,c)$ since you can remove four adjacent vertices with one of them being white. Using this repeatedly, we have $(32,32,15)\implies$ $(59,41,24).$ We claim that anything of the form $(2n,2n,n-1)$ is true. We can kick out a vertex to get $(2n,2n-1,n-1)$, and \[(2n-3,2n-2,n-2)\equiv (2n-2,2n-3,n-2)\implies (2n,2n-1,n-1)\]so by induction we are done.

Claim: If there are $w$ white vertices and $b$ black vertices, where $w>b$, then there are $4$ consecutive vertices consisting of $3$ white and $1$ black. Proof: Label the vertices $1, \ldots, w+b$, and let $S_i$ be $1$ if vertex $i$ is white and $0$ if vertex $i$ is black, where indices cycle. We make a more general claim that there are $4$ consecutive vertices consisting of $3$ white and $1$ black or $4$ white and no black. Assume for contradiction that this is not the case. Then, $S_i+S_{i+1}+S_{i+2}+S_{i+3} \le 2$. However, \[ 2(w+b) \ge \sum_{i} S_i+S_{i+1}+S_{i+2}+S_{i+3} = 4\sum_{i} S_i = 4w, \]which yields $w \le b$, a contradiction. Thus, there are $4$ consecutive vertices consisting of $3$ white and $1$ black or $4$ white and no black. If this selection returns $4$ white and no black, then shift until a $WWWB$ pattern is obtained. This proves the claim. $\square$ Now, delete one white vertex, and repeatedly delete quadrilaterals that exist per the claim. Then, the distribution (# of white vertices, # of black vertices) becomes \begin{align*} (58,41) &\to (55, 40) \to (52, 39) \to (49, 38) \to (46, 37) \to (43,36) \\ &\to (40,35)\to (37,34)\to (34,33) \to (31, 32) \to (30,29) \\ &\to (27,28) \to (26,25)\to (23,24) \to (22, 21) \to (19,20)\\ &\to (18,17) \to (15,16) \to (14,13) \to (11,12) \to (10,9) \\ &\to (7,8) \to (6,5) \to (3,4) \to (2,1) \end{align*}with each deletion. This contributes $24$ deletions, which corresponds to $24$ quadrilaterals, and we are done.

Suppose there are $b$ black vertices and $w$ white vertices. If all vertices are black or white, we are done. I claim that when $b\ne w$ and both are positive, we can find a quadrilateral with four consecutive vertices, three of which are same-colored with the last being different. WLOG $b>w$. I claim that we can find a quadrilateral with three black vertices and one white vertex. Suppose not; then notice that a block $BBBB$ cannot occur either, since we must obtain $BBBW$ or $WBBB$ on the edge of the $B$-run containing this block. Hence all blocks contain at least as many $W$ as $B$. Summing globally, we find $w\ge b$, a contradiction. Notice that, more specifically, the only way such a quadrilateral does not exist is if every block contains two $B$ and two $W$. We can clearly apply the algorithm continuously until such a state indeed occurs. Observe that we must find $n$ quadrilaterals among a $4n+4$-gon. Verify that the polygon has runs of uniform size either $1$ or $2$ (either $BWBW$ or $BBWW$). If the polygon has runs of length $2$, then split into two equal-length segments as follows (we do need to split by $\pmod 8$ but the other case should be analogous): \[BBWW\dots BBWW\]\[WWBB\dots WWBB\]Truncate (1) the $WW$ at the end of the first segment and (2) the $W$ and $B$ at the ends of the second segment. Now we have \[BBWW\dots BBWWBB\]\[WBBW\dots WBBWWB\]and verify that we can pair adjacent columns to obtain the desired quadrilaterals. In particular we have removed $4$ vertices and so this is fine. If the polygon has runs of length $1$, then the following suffices: WLOG vertex $1$ is white, take \[(4n+2,4n+3,4n+4,2)\]\[(4n-1,4n,4n+1,3)\]\[\vdots\]\[(n+5,n+6,n+7,n+1)\]of which there are $n$ quadrilaterals in total.

what First, assume we have some generic polygon where WLOG $w>b\ge 1$. There exists some block of four consecutive vertices where three or more vertices are white by Pigeonhole. Then if a block of four white vertices exists then as one black vertex exists we know there is a block of exactly three white vertices. Ignore a single white vertex so that there are an even number of white vertices and an odd number of black vertices. Now run the process; clearly the number of white and black vertices is always different. In particular we can directly track the count of vertices: \[(41,58)\to (40,55)\to (39,52)\to (38,49)\to (37,46)\to (36,43)\to (35,40)\to (34,37)\to (33,34)\to (32, 31)\to (29,30)\to (28,27)\to \dots\to (4,3)\to (1,2)\]and we are done. $\blacksquare$

Pretty silly problem. We will define a set $S$ of ``considered" points where $S$ starts out as all $100$ points, and we delete points from $S$ either by choice or by removing a quadrilateral. Claim. If $S$ contains more black points than white points, then there exists four consecutive points in $S$ with exactly three of them black. Proof. Observe that for double counting reasons, there must exist four consecutive points in $S$ that contain at least three black points. If all four points are black, then we shift the set to the left or right arbitrarily; we have to run into a white point at some point, creating the desired set. $\blacksquare$ Now we will pick such a consecutive set of $3$ white and $1$ black points from $S$, create a quadrilateral $Q_1$, and remove them from $S$. This leaves us with $40$ black and $56$ white points. Using this claim again, we can create nine white-majority quadrilaterals $Q_1, Q_2, \dots, Q_9$, leaving $S$ consisting of $32$ black and $32$ white points. At this point, we may delete a white point from $S$ and create quadrilaterals $Q_{10}, Q_{11}, \dots$ from consecutive groups with alternating majority colors. This always works because the majority color alternates between black and white, and the number of black and white points are never equal. (To illustrate, the number of (black, white) points goes $(32, 31) \to (29, 30) \to (28, 27) \to \cdots \to (4, 3) \to (1, 2)$.) This yields $15$ more quadrilaterals $Q_{10}, Q_{11}, \dots, Q_{24}$, as needed.

We form quadrilaterals by taking 4 consecutive points on a circle. Claim: When there are more white points than black points (and both colors exist), we can find 4 in a row such that three are white, one is black. Proof: It is obvious that we can choose some four adjacent points where there are more white than black (sum over every adjacent quadruple). Then the only possibility of there not being a 3-1 split is when it's a 4-0 split, contradicting the existence of black points. So therefore, we can do this to find a quadrilateral that is "good". By geometry, if we only choose quadrilaterals with vertices adjacent on the circumference, they will always be disjoint. So now, we can choose "good" quadrilaterals until we have 32 white and black points respectively. Now we choose five consecutive points along the circumference that are split 3-2 in either way. WLOG we have 3 blacks, 2 whites. Ignore one of the whites, and draw the quadrilateral with the other four points as vertices. We impose a restriction that prevents our ignored point from being a vertex of any quadrilateral. Now, we have 30 white points, 29 black points. Claim: When we have $|W-B|=1$ we can keep choosing "good" quadrilaterals while preserving the $|W-B|$ quantity, until we reach $B=2,W=1$ or vice versa. Proof: This is a simple induction proof. Skipped because I am not bothered This basically allows us to choose quadrilaterals until we have 2 points of one color and one of the other remaining. Therefore, we have 4 points (remember the ignored vertex) that are not in a quadrilateral, so the other 96 make up exactly 24 quadrilaterals that satisfy the necessary conditions. The end.