I assume that a group of more than $100$ students involves counting certain students twice. That is kind of obvious, but nevertheless it is pretty poorly stated. A contiguous group of between $200$ and $600$ students out of $100$ students doesn't really exist. Uh but $k=100$ just gives the first half of the group of $200$ students is... all the students, and so is the last half, so $k=100$ satisfies the problem and this is trivial... Should $100$ be $1000$ or some other bigger number?

There are $1000$ students. Typo perhaps, but seeing that OP didn't edit his post... Number the students in order from 1 to 1000. Let $a_i = 1$ if student $i$ is a girl, and $a_i = 0$ otherwise. Also have indices modulo 1000. Let $f(n,k) = \displaystyle\sum_{i=n}^{n+k-1} a_i - \displaystyle\sum_{i=n+k}^{n+2k-1} a_i$ for all integer $n$ and integer $k \in [100,300]$. If we have $f(n,k) = 0$ for some $n,k$, then we are done. Suppose otherwise. It's pretty obvious that $f(n+1,k) - f(n,k) \in [-2, 2]$, since $f(n+1,k) - f(n,k) = a_{n+2k} + a_n - 2a_{n+k}$. Then, $\displaystyle\sum_{i=1}^{1000} f(i,k) = 0$ for all $k$, because each student is counted for $k$ times as positive and $k$ times as negative. Hence there must be some $n$ where $f(n,100) > 0$ and $f(n+1,100) < 0$; otherwise the sequence $f(i,100)$ is either all positive or all negative, impossible. Slight analysis gives $f(n,100) = 1$ and $f(n+1,100) = -1$ because otherwise $f(n+1,100) - f(n,100) > 2$. Hence we have the sequence $BXGXB$ where $B$ indicates boy (or non-girl, as the problem didn't specify there is only one other gender ), $G$ indicates girl, and $X$ indicates a sequence of 99 students where the number of girls between the two groups are equal. Now take $f(n-1,101)$. If $a_{n-1} = 1$ then $f(n-1,101) = 0$, contradiction. Hence $a_{n-1} = 0$. Similarly, $a_{n+201} = 0$ from $f(n,101)$. Inducting, we have $a_{n-k} = 0$ for all $k \in [0, 200]$. But then $f(n-200, 100) = 0$, contradiction.

chaotic_iak wrote: but seeing that OP didn't edit his post When moderators edit posts in the forums they moderate, it doesn't list the edits.

Yes, it was 100 at first, and I PM'ed the OP. He then edited it.

I have almost the same solution to chaotic_iak,nice problem. Assume contrary. Now,let $a1,a2...a1000$ be numbers that are $1$ or $0$.Now,let $b1,b2,...b1000$ be places beetwen $a1a2,a2a3,....a1000a1$.Now,if we pick $200$ sequences that have the same "beetwen" place,we easy obtain every sum clockwise has more $1s$ than the sums in the opposite direction,cause if not we will have equal two equal sums.Now,for every "beetwen" place denote the direction of bigger sums.Now,suppose that all "beetwen" places have the same direction,but that is not true if we divide $1000$ students in $10$ consecutive groups,so we conclude that we must have two consecutive "beetwen" places that have opposite directions,so we conclude that we will have a sequence like this WLOG $1X0X1$,where $X$ denotes a group of $99$ students that have the same number of $1s$.Now,we easy obtain that in both directions we have $200$ consecutive $1s$,so we are done.

Assume contradiction. Consider the following structure: A block of k=100 students with X girls followed by 100 students with Y girls. As we rotate this structure around the circle, Y-X can never be 0. But notice that when we rotate it 1 time, its value changes by d, where d=-2,-1,0,1 or 2. It is easy to see that sometimes Y-X<0 and sometimes Y-X>0 (add all the 1000 values of Y-X and the sum will be 0). Hence, when it crosses 0, it goes from Y-X=1 to Y-X=-1. When this happens, there is a boy, 99 students in between (these students form X), a girl, 99 students in between (these students form Y) and then a boy. Something like this: B, x,...,x, G, y,...,y, B (where the first boy is in X and the girl is in Y, and there are 99 xs and ys). Now consider k=101. And expand the group Y to include the second B and the group X to include the student before the first B. This proves that the student before the first B is a boy. Now expand the original group X to include G and shift the original group Y to include the second B and the student after him, but no longer include B. This proves the student after the second B is also a boy. B, B, x, ..., x, G, y, ..., y, B, B Repeat the argument with k=102,...,300 and you get the structure B,...,B, x, ..., x, G, y, ..., y, B, ..., B where there are 200 boys in the first block. Then with k=100 consider this block of 2k students. The first k are all boys and the second k are all boys, and thus we have found a set. QED Comment: Let a k such that such a set exists be "bad". Then I proved that in any interval [x,3x], there is a bad k. Can we get a good bound on the number of bad k's in (1,500), using this?

Assume for contradiction there doesn't exist such a $k$. Consider $k = 100$. For each of the contiguous group of $200$ students, write a number that is the difference between the number of girls of first $100$ and that of the second $100$. Each time shifting the $200$ students cw by $1$, the number changes by at most $2$. Furthermore, each girl is adds $1$ to $100$ of these numbers and causes a subtracting of $1$ in $100$ of these numbers so that the sum of all the written numbers is $0$. Start with one such group and wlog the corresponding number is positive then there exists another group that has a corresponding negative number. So we must have "crossed" $0$ as we move towards that group. No number is $0$ so by prev observation there exists a step where we go from a $1$ to a $-1$. So along the circle we have something like \[(0)(A)(1)(B)(0)\]where $A$ and $B$ are blocks of $99$ people with the same number of girls. Consider the two groups $(0)(A)$ and $(1)(B)(0)$, the latter has $1$ more girls so the person to the left must be $0$ (otherwise there exists a good $k$). Similarly consider the two groups $(0)(A)(1)$ and $(B)(0)$ so now we have: \[(0)(0)(A)(1)(B)(0)(0)\] Eventually before considering two groups involving sizes more than $301$ we get that there is a contiguous string of $200$ zeros which is a contradiction.

Replace boys with $0$ and girls with $1$. We just want to find to adjacent blocks of length $k$ with the same sum. Suppose the problem statement is false. Suppose we have two blocks of length $k$ of the form $(x\cdots)(z\cdots)w$ (blocks are denoted in parantheses, the rest is the surrounding numbers). If we shift our blocks by $1$ to the right, we see that the net change in $S$ (the differnce of the right block sum and the left block sum) is $w-z-(-x+z)=w+x-2z$. In particular, unless $w=0,x=0,z=1$, we have that the change in $S$ is at least $-1$, and in the mentioned case it is $-2$. Split the $1000$ students into $10$ blocks of size $100$. Going clockwise, label the block sums $A_1,A_2,\ldots,A_{10},A_{11}=A_1$. We must have $A_i<A_{i+1}$ and $A_j> A_{j+1}$ for $j>i$. Thus, as we slide blocks $A_i,A_{i+1}$ into $A_j,A_{j+1}$, we have that the quantity $S$ goes from positive to negative. But as it never achieves $0$ (we are assuming the problem statement is false), we must have that it goes from $1$ to $-1$ since $\Delta S\ge -2$ always. In this case, $w=0,x=0,z=1$ from the above notation, so we have two consecutive blocks of $100$ of the form \[(0A)(1B)0\]where $A$ and $B$ have the same sum. Let's now ignore the block structure, so we know we have a portion of the circle that looks like $0A1B0$ where $A$ and $B$ are strings of length $99$ with the same sum. We now claim that we must have in fact at least $200$ zeroes on each sides of $A1B$ (we can probably get it to $298$, but $200$ suffices). If not, then consider the $1$ that is closest to the center $1$ (if there are two, pick one arbitrarily). If we have then \[0^{k-1}A1B0^{k-1}1,\]then we clearly have a contradiction by picking $(0^{k-1}A1)(B0^{k-1}1)$. Thus, we have a contiguous block of $200$ zeroes, so we have a contradiction, as desired. $\blacksquare$

This is a funny problem Let the students have a value of one if they are a girl, and a value of zero otherwise. (The sum of all elements in a contiguous block of students just counts the number of girls in that block.) Label the students $1$ to $1000$ in, say, clockwise order, with indices taken modulo $1000.$ Suppose the result weren't true; That is, for any $100 \leq k \leq 300,$ and any contiguous block of $2k$ students, both of the two halves of $k$ students have the same sum. Let $S(a,b)$ denote the sum of the students' values starting from student $a$ and ending at student $b,$ going clockwise. Now consider $k=203.$ Note that $\gcd(203, 1000) = 1.$ Starting with student 1, we then have $S(1, 203) = S(204, 406) = S(407, 609) = \cdots$ Since 203 is relatively prime to 1000, this means all 1000 of these contiguous sums of length 203 have the same value! Now consider $k=202.$ Note that $\gcd(202,1000)=2.$ Then we must have $S(1,202)=S(203,404)=S(405,606) = \cdots,$ $S(2,203)=S(204,405)=S(406,607) = \cdots,$ so by gcd we know the all sums starting on even indices are equal, and likewise for odd indices. In summary,we have: (i) $S(2n, 2n+201)$ is constant for all $n,$ (ii) $S(2n+1, 2n+1+201)$ is constant for all $n.$ (iii) $S(m, m+202$) is constant for all $m.$ however this means $val(203)=val(405)=val(607) = \cdots$ and $val(204)=val(406)=val(608)=\cdots,$ by subtracting (i) and (ii) from (iii). In other words, all even indices have the same value, and all odd indices have the same value. There are four cases, and each can be verified clearly to satisfy the problem condition: If all students have same value, or if the students alternate values, both satisfy the problem condition. This is a contradiction. _______________________________________________________________________________________________________________ Comment: The really funny part about this problem is that the condition $100 \leq k \leq 300$ is quite unnecessary; you only need to find a few specific values of $k,$ and it all goes down from there. Of course, extremal arguments with contradiction are probably better solutions.

Interesting problem! Here's my solution: Suppose the problem statement is not true. Label the students as $1,2, \dots ,1000$ in clockwise fashion. Note that there are a total of $1000$ contiguous blocks of length $200$. Define the measure $m(i)$ of an integer $i \in [1,1000]$ as the number of girls in the first block of $100$ students minus number of girls in the next block of $100$ students, as we start from the $i^{\text{th}}$ student, and move clockwise. Also, for the $i^{\text{th}}$ student, let the indicator of that student be such that this value is $1$ if that student is a girl, and $0$ otherwise. For some $i \in [1,1000]$, let $a,b,c$ be the indicators of the students labelled $i,i+100,i+200$ (all labellings taken modulo $1000$). Then we easily get $$m(i+1)-m(i)=2b-a-c \in \{-2,-1,0,1,2\} \quad (\heartsuit)$$Also, we have $$m(1)+m(2)+ \dots +m(1000)=0 \quad (\spadesuit)$$since here the number of girls in each contiguous block of length $100$ are once added, and once subtracted. Since the problem statement is untrue, so none of the measures are $0$. Then, from $(\spadesuit)$, not all $1000$ measures can have the same sign, and so there must exist some $i \in [1,1000]$ such that $m(i)>0$ and $m(i+1)<0$. Since the measures are integers, so we get $$m(i) \geq 1 \text{ and } m(i+1) \leq -1 \Rightarrow m(i+1)-m(i) \leq -2 \Rightarrow \text{ Using } (\heartsuit), m(i+1)-m(i)=-2$$Letting $a,b,c$ denote the indicators as before, this is only possible if $b=0,a=1,c=1$. So that means that we have a block of $201$ students with the following form $$G_1 \underbrace{\dots \dots \dots}_{99 \text{ students}} B \underbrace{\dots \dots \dots}_{99 \text{ students}} G_2$$where the groups of $99$ students (say $S_1$ and $S_2$) have an equal number of girls (Here $G_1,G_2$ are girls, and $B$ is a boy). Now, if the next student after $G_2$ is a boy $B'$, then $G_1S_1B$ and $S_2G_2B'$ give two consecutive sets of $101$ students each, having an equal number of girls. Then we will have our desired contiguous block of length $202$, a contradiction. So the next student after $G_2$ must be a girl $G_2'$. Similarly, the previous student before $G_1$ must also be a girl $G_1'$. Similarly, if the next student after $G_2'$ is a boy $B'$, then the blocks $G_1'G_1S_1B$ and $S_2G_2G_2'B'$ contradict our original assumption. Continuing in this manner, we get that the next $200$ students after $G_2$, and the previous $200$ students before $G_1$ must all be girls (since $1000>200+200+201$, so all these students are distinct). But then any of these sets of $200$ girls give our desired contiguous group. Hence, done. $\blacksquare$

Let $0$ and $1$ denote boy and girl, respectively. For contradiction, assume that no such $k$ exists. Consider two adjacent blocks and the letter right after, say $[a\ldots][b\ldots]c$. Let $S$ denote the right block sum minus the left block sum. Note that as we shift both blocks around the circle by $1$ to include $c$, the value of $S$ experiences a net change of $(c - b) - (b - a) = a + c - 2b$. Notice that since $a, b, c \in \{0, 1\}$, in fact plugging in all possible values, we get that $a + c - 2b \geq -1$ and it equals $-2$ iff $(a, b, c) = (0, 1, 0)$. Now we choose some two blocks, each of size $100$, such that $S$ is negative. There must also exist some two blocks, each of size $100$, such that $S$ is positive. As we shift the "positive" blocks toward the "negative" blocks, $S$ must change from positive to negative, where the only possible jump is $1 \to -1$ since $S$ changes by at most $-2$ each time. So at this point, we have some two adjacent $100$-size blocks arranged in the following format:\[0[a_1a_2\ldots a_{99}]1[b_1b_2\ldots b_{99}]0\]where the blocks are $0[a_1a_2\ldots a_{99}]$ and $1[b_1b_2\ldots b_{99}]$. Note that the sum of $a_i's$ must be equal to the sum of $b_i's$ since before the jump, $S = 1$. Denote the block of $a_i$ as $A$ and $b_i$ as $B$ for simplicity, and these two blocks must have equal sum. I claim that the $200$ terms before and after $0[A]1[B]0$ must all be zeroes. Consider the following inductive argument. Suppose that at some point, we have\[\underbrace{0\ldots0}_{\text{k zeroes}}[A]1[B]\underbrace{0\ldots 0}_{\text{k zeroes}}\]and if we consider block of $k + 99$ going left from $A$ and the block of $k + 100$ going right from $1$, we see that if we want to add a term to the left of the leftmost $0$ and still keep the two aforementioned blocks from having equal sum, this added left term must be $0$. We do this again but switch the roles of left and right, so the new added rightmost term must also be $0$. Eventually, we get to a stage where there are $200$ zeroes on both sides, and we cannot continue the process any further because $1[b]\underbrace{0\ldots0}_{\text{200 zeroes}}$ has size $300$. However, at this point, our problem is finished, as since we have a block of $200$ consecutive zeroes, we break it into two consecutive blocks of $100$ zeroes each, which have equal sum, a contradiction, as desired. $\blacksquare$ Remark: This is a great problem.

Solved with nukelauncher. Let \(a_i=1\) if the \(i\)th student is a girl and \(a_i=0\) if the \(i\)th student is a boy (with indices taken mod 1000); then consider \[f(n,k)=\sum_{i=n}^{n+k-1}a_i-\sum_{i=n-k}^{n-1}a_i.\] Observe that \begin{align*} |f(n+1,k)-f(n,k)|&\le2\\ |f(n,k+1)-f(n,k)|&\le1 \end{align*}for all \(n\), \(k\). Assume for contradiction that \(f(n,k)\ne0\) for all \(n\) and \(100\le k\le300\). Claim: For each \(n\), all \(f(n,k)\) have the same sign for \(100\le k\le300\). Proof. Else, by \(|f(n,k+1)-f(n,k)|\le1\), some \(f(n,k)\) would equal zero. \(\blacksquare\) Claim: For each \(k\), there must exist \(n\) with \(f(n,k)=+1\) and \(f(n+1,k)=-1\). Proof. Since \(\sum_nf(n,k)=0\), we would otherwise find some \(f(n,k)\) equal to zero. \(\blacksquare\) Without loss of generality, let \(f(0,k)=+1\) and \(f(1,k)=-1\). Claim: \(f(0,\ell)=+1\) and \(f(1,\ell)=-1\) for all \(100\le\ell\le300\). Proof. By the first claim, \(f(0,\ell)>0\) and \(f(1,\ell)<0\); however, recall that \[|f(1,\ell)-f(0,\ell)|\le2.\]\(\blacksquare\) Applying the third claim, we find the following: since \(f(0,\ell)=+1\;\forall\ell\), we have \(a_{100}=a_{-101}=a_{102}=a_{-103}=\cdots=a_{-299}=a_{300}\); and since \(f(1,\ell)=-1\;\forall\ell\), we have \(a_{-100}=a_{101}=a_{-102}=a_{103}=\cdots=a_{299}=a_{-300}\). Finally, \begin{align*} f(-200,100)&=(a_{-200}+\cdots+a_{-101})-(a_{-201}+\cdots+a_{-300})\\ &=(a_{-200}-a_{-300})+(a_{-199}+-a_{-299})+\cdots+(a_{-101}-a_{-201})\\ &=0, \end{align*}contradiction.

Replace circle of students by infinite periodic sequence of numbers $$a_i = \begin{cases} \ 1 & \text{if it corresponds to a girl,} \\ 0 & \text{otherwise.} \end{cases}$$Define $\sigma (n,k)=\sum_{i=n+k}^{n+2k-1} a_i-\sum_{j=1}^{n+k-1} a_j.$ We have to prove that $\exists n, 100\leq k\leq 300: \sigma (n,k)=0.$ Assume the opposite. Claim 1. There exists such $m$ that $\sigma (m,100)=1$ and $\sigma (m+1,100)=-1$. Proof. Together with our asumption it suffices to observe the following facts: $|\sigma (n+1,k)-\sigma (n,k)|\leq 2$ by a straightforward check; $\sum_{i=1}^{1000} \sigma (n,100)=0$ since all $a_i$ are considered $100$ times with both positive and negative signs in this sum. $\Box$ Claim 2. For every $0\leq l\leq 200$ holds $a_{m-l}=0=a_{m+200+l}.$ Proof. Induct on $l;$ by the previous claim starting from $a_m$ we have a subsequence $$0A1B0,$$where $A,B$ denote two sequences of $99$ zeroes and ones with the same sum; in particular, this proves the base case $l=0$. Suppose that starting from $a_{m-l}$ we have a subsequence $$\underbrace{00\dots 0}_{l+1\text{ zeroes}}A1B\underbrace{00\dots 0}_{l+1 \text{ zeroes}}.$$Then $\sigma (m-l-1,100+l)\neq 0$ yields $a_{m-l-1}=0,$ and similarly $a_{m+201+l}=0,$ completing the inductive step $\Box$ However, due to the second claim we deduce that $\sigma (n-200,100)=0,$ which is the desired contradiction.

Very nice problem!! Replace boys and girls by 0 and 1. We consider the blocks of 100 first: Consider the difference between the values of two consecutive blocks b_1,b_2, where b_2 is directly clockwise of b_1, where a block's value is the number of girls in there, and let these differences be d_n which is a sequence. FTSOC this value is never 0; note that because the total sum is 0 (each girl is considered in the difference of the block in counterclockwise and clockwise equal number of times), there must be some negative value, and since |d_i-d_{i+1}| is at most 2 since only two girls' are added/subtracted from the difference. Then, since the total sum is 0, there must be positive and negative values, so it must shift from 1 to -1 WLOG. This means that there must be strings of 99 people 0S_11S_20, with |S_1|=|S_2|; considering k=101 with 1S_20 vs. 0S_1, this differs by 1, whence 0 must be to the left of 0S_1. Similarly, we get 0 is the right of S_20; continuing this way until k=300 we have 201 consecutive zeros, and now there are indeed two consecutive blocks, each with value 0, so that difference is 0, as desired. $\blacksquare$

Suppose for the sake of contradiction that no such group exists. Define a split point as a space between two people in a circle. For a given split point $s$, let $$D_{s,k}$$denote (number of girls in the $k$ students immediately counterclockwise of $s$) minus (number of girls in the $k$ students immediately clockwise of $s$). The hypothesis is that for any split point $s$, $D_{s,100}$ through $D_{s,300}$ are all nonzero. Note that however, if we increase $k$ by 1, then $D_{s,k}$ changes by at most 1. Since it never becomes zero, we can conclude that for any given $s$, we have that $$D_{s,100}$$all the way through $$D_{s,300}$$are all the same sign. We say that a split point $s$ is positive if all of these $D_{s,k}$ for $100\leq k\leq 300$ are positive, and negative if all of these are negative. WLOG one of the split points is positive since otherwise we can just reflect. Claim: If a split point $s$ is positive, then the one immediately clockwise of it, which we call $s+1$, is also positive. Assume FTSOC that it is negative. We will first show a sub-claim: Sub-claim: $$|D_{s,k}-D_{s+1,k}|\leq 2.$$When shifting the split point clockwise by 1, a student furthest counterclockwise is lost, a student furthest clockwise is lost (goes from zero to negative weight), but the student that was shifted over is gained twice (goes from negative to positive), which clearly shows this sub-claim. Now, this means that the only way for $s+1$ to be negative is if for all $100\leq k\leq 300$ we had $$D_{s,k}=1,D_{s+1,k}=-1.$$Furthermore, consider the sub-claim again. The only way we could lose 2 when going from $D_{s,k}$ to $D_{s+1,k}$ is if the two students that were lost were both girls, but the one that was gained twice was a boy. However, this must apply for all $100\leq k\leq 300$. This means that this produces (actually two) sequences of at least $200$ consecutive girls, as we reach one student further for each new value of $k$, which clearly contradicts our assumption directly. Going all the way around the circle with our claim, all split points are positive. However, we clearly have $$\sum_{s}D_{s,100}=0$$as each student is counted positively and negatively an equal number of times as we sum all split points. However, this is not possible if all split points are positive, contradiction.

Very beautiful but very tricky. Label the positions in the circle as $a_1, a_2, \dots, a_{1000}$, where $a_i= 0$ if $i$ is a boy and $a_i = 1$ if $i$ is a girl. We define the mensity difference function $d(n, k)$ to equal $a_{n+1}+a_{n+2} +\cdots+a_{n+k} - (a_n+a_{n-1}+\cdots+a_{n-k+1})$ for each pair $(n, k)$. The problem asks us to show that there exists an $n$ such that $d(n, k) = 0$ for some $100 \leq k \leq 300$. To show this, notice that $|d(n, k) - d(n, k+1)| \leq 1$ and $|d(n, k) - d(n+1, k)| \leq 2$ for all $n, k$. So, by a quasi-IVT argument (!!) it suffices to show the following: There exists $n_1$ such that $d(n_1, 100) < 0$; There exists $n_2$ such that $d(n_2, 300) > 0$; There exists a sequence of transformations shifting $n$ or $k$ by $1$ at every step that transforms $d(n_1, 100)$ to $d(n_2, 300)$ and passes through some $d(n, k) = 0$. The first two conditions are obviously true; to see this, compare the sum $S(n)$ of the $100$ terms starting with $a_n$ and note that $\{S(n), S(n+100), \dots, S(n+900)\}$ has a unique minimum. By the original two observations, there will always exist $d(n, k) = 0$ during the sequence of transformations from $d(n_1, 100)$ to $d(n_2, 300)$ unless there exists an $n_0$ such that $d(n_0, k) = -1$ and $d(n_0+1, k) = 1$ are consecutive terms in the sequence. Suppose that this indeed happens. Then, it follows that $a_{n_0-k + 1} = 1$ (as it is the subtracted term ``lost" during the transformation) and $a_{n_0+k + 2} = 1$ (as it is the added term ``gained" during the transformation). But observe: We have $d(n_0, k+1) \geq d(n_0, k)$ as $a_{n_0+k+2} = 1$ and thus $d(n_0, k+1) - d(n_0, k) \in \{0, 1\}$; We have $d(n_0, k-1) \leq d(n_0, k)$ as $a_{n_0-k+1} = 1$ and thus $d(n_0, k-1) - d(n_0, k) \in \{-1, 0\}$. Replacing $k$ with $k' = k-1$ or $k+1$, we may iterate this repeatedly a total of $k - 100$ times down and $300-k$ times up while keeping $100 \leq k' \leq 300$. We look at the upward transformations WLOG. Within these iterations, either there exists some $d(n_0, k'+1) - d(n_0, k') = 1$ which will yield $d(n_0, k'+1) = 0$ satisfying the problem statement, or $d(n_0, k'+1) = d(n_0, k')$ everywhere, implying that $a_{n_0-k'} = 1$ for all $k \leq k' \leq 300$. Similarly, either some $d(n_0, k'-1) = 0$ for some $100 \leq k' \leq k$ or $a_{n_0 - k'} = 1$ for all those $k'$ too. Thus, the only case remaining is when $a_{n_0-k'} = 1$ for all $100 \leq k' \leq 300$. But then we have $200$ consecutive $a_i$ with $a_i = 1$ and the result is now clear.

Assume that the problem statement is wrong for the sake of contradiction. We'll rewrite all girls by $1$ and all boys by $0$. Let $a_1,a_2,...,a_{1000}$ be the numbers written contiguously on the circumference. We'll now consider the indices to be taken modulo $1000$. Now, define $b_i=a_{i+100}-a_i$ for all integers $i$. We note that $$\sum_{i=1}^{1000} b_i =0$$Thus, $$\sum_{j=1}^{10} \left( \sum_{i=100j-99}^{100j} b_i \right)=0$$We now conclude that there's exist $l\in \mathbb{Z}$ such that $$\sum_{i=l}^{l+99} b_i<0, \text{but} \sum_{i=l+100}^{l+199} b_i>0$$Now, define the sequence $$c_s=\sum_{i=l+s}^{l+s+99} b_i$$for all $s\in[0,100]$. We now have $c_0<0$, $c_{100}>0$, and $c_{s+1}=c_s+b_{l+s+100}-b_{l+s}\in \{c_s+2,c_s+1,c_s,c_s-1,c_s-2 \}$ Hence, we can ensure that there's some index $m$ such that $$\sum_{i=m}^{m+99} b_i=-1, \text{and} \sum_{i=m+1}^{m+100} b_i=1$$Thus, $-1+b_{m+100}-b_m=1 \implies b_{m+100}=1, \text{and } b_m=-1$. Now we can see that $$\sum_{i=m+1}^{m+99} b_i=0 \iff \sum_{i=m+1}^{m+99} a_i=\sum_{i=m+101}^{m+199} a_i$$WLOG $a_{m+100}=1$ (the other case can be checked similarly). If $a_{m+200}=1$, then $$\sum_{i=m+1}^{m+100} a_i=\sum_{i=m+101}^{m+200} a_i$$, a contradiction. If $a_m=1$, then $$\sum_{i=m}^{m+99} a_i=\sum_{i=m+100}^{m+199} a_i$$, a contradiction. Hence, we now assume $a_m=a_{m+200}=0$. If, $a_{m-1}=1$, then $$\sum_{i=m-1}^{m+99} a_i=\sum_{i=m+100}^{m+200} a_i$$, a contradiction. If $a_{m+201}=1$, then $$\sum_{i=m}^{m+100} a_i=\sum_{i=m+101}^{m+201} a_i$$, a contradiction. Hence, we now assume $a_{m-1}=a_{m+201}=0$. We continue this process until we get that $$\sum_{i=m-200}^{m+99} a_i=\sum_{i=m+101}^{m+400} a_i$$with the facts that $a_{m+200}=a_{m+201}=...=a_{m+400}=0$. Now, we see that $a_{201},a_{202},...,a_{400}$ contradicts our first assumption. Hence, there must exist $k\in[100,300]$ that satisfies the problem condition.

Let $a_i=1$ if the $i\pmod{1000}$th person is a girl and $0$ otherwise. Assume for the sake of contradiction that for all $100\le k\le 300$ and $i$, the value \[f_k(i)=\sum_{j=i-k+1}^{i}{a_j}-\sum_{j=i+1}^{i+k}{a_j}\]is nonzero. Note that $f_k(1)+f_k(2)+\dots+f_k(1000)=0$ and $f_k(i)-f_k(i-1)\le 2$ with equality if and only if $a_{i-k}=a_{i+k}=0$ and $a_{i}=1$, so there exists $i$ such that $f_{100}(i-1)<0<f_{100}(i)$ so $f_{100}(i-1)=-1$ and $f_{100}(i)=1$. This implies $a_{i-100}=a_{i+100}=0$ and $a_i=1$. Furthermore, we know that \[\sum_{j=i-99}^{i-1}{a_j}=\sum_{j=i+1}^{i+99}{a_j}\]Let $r$ be the smallest number greater than or equal to $100$ such that $a_{i+r}$ or $a_{i-r}$ is equal to $1$. Then either $f_r(i-1)$ or $f_r(i)$ is zero. Thus, $r\ge 301$. However, this implies that $a_{i+100},a_{i+101},\dots,a_{i+300}$ are all zero, which means $f_{100}(i+200)=0$, contradiction.

For any $200$ students, define their "difference" as \[ (\text{\# of girls in the clockwise-most } 100) - (\text{\# of girls in the counterclockwise-most } 100).\] If we choose any $200$ students at random, by symmetry, their difference has an average value of $0$. Furthermore, the difference changes by at most $2$ when we shift our block one student clockwise. If there exists a block of students whose difference is $0$, we are finished. Otherwise, there must be a block $B$ of $201$ students for which the counterclockwise $200$ students have a difference of $1$ and the clockwise $200$ students have a difference of $-1$. (Or vice versa, but we don't have to consider that because it's symmetric to this case by toggling the gender of each student.) From our definition of $B$, it follows that the first and last students in $B$ are not girls, and that the $101$st student in $B$ is a girl. Call the $101$st girl Chloe. If there exists another girl $\mathcal X$ outside of block $B$, such that there are at most $298$ students in between Chloe and $\mathcal X$, pick the closest such $\mathcal X$ to Chloe – ties don't matter. WLOG, assume that $\mathcal X$ is counterclockwise of Chloe. Let $k$ be one plus the number of students strictly between Chloe and $\mathcal X$, so that $101 \leq k \leq 299$. We choose our block $C$ of $2k$ students such that $\mathcal X$ is the counterclockwise-most student in the block. It is easy to see that $C$ meets the requirement of an equal number of girls in each half. If no such girl exists, we can select the clockwise most student (call him Stanley) of $B$ plus the $199$ students clockwise of Stanley, none of which are girls. Then, we just choose these $200$ students and we are done, since $0=0$.

Everyone is saying its great but my solution is just sad. Its near trivial by probablistic