Well, if $in$ has residue $r_i$ mod $202$, then $\sum r_i = 101k$. But $\sum r_i \equiv n\frac {k(k+1)}{2} (mod 202)$, so $101 | n\frac {k(k+1)}{2}$. If $101$ doesn't divide $\frac {k(k+1)}{2}$, then $101 |n$ but this obviously does not work except for $k=1$, so $k=100, 101, 201$. It is left to add examples for such $n$ for all those $k$, which I might try to do later. Edit: for $k=100$ take $n=2$; for $k=201$ take $n=1$. Those are motivated by grouping terms and looking for the avarage to be $101$

Redacted

I don't think n=3 works at k=101.

The answer is $k=1,100,101,201$. For $k=1,n=101$ works. For $k=100,n=2$ works For $k=101,n=103$ or $n=51$ works. ($n=51$ works is a bit complicated) For $k=201,n=1$ works. For all other $k$, note $101\nmid k(k+1)$ but $\frac{k(k+1)n}{2\cdot 202} \equiv \frac k2 (\bmod\; 1)$. This means $n\equiv 0$ or $n\equiv 101 (\bmod\; 202)$. Clearly $202\nmid n$. If $n\equiv 101(\bmod\; 202)$ then we note LHS is half the number of odd numbers in $[1,k]$. Thus, $k=1$, the end.

Just wanted to ask: what's the motivation for finding a value of $n$ that works for $k=101$? I spent the entire time convinced that nothing worked Python gives me loads of weird solutions, how do we find those?

how many points just providing the correct answer set is worth?

InternetPerson10 wrote: Just wanted to ask: what's the motivation for finding a value of $n$ that works for $k=101$? I spent the entire time convinced that nothing worked Python gives me loads of weird solutions, how do we find those? You just have to believe -- that's how I got it.

Note that $202\bigg\{\frac{a}{202}\bigg\}$ is the remainder of $a$ when divided by $202$, therefore $202\bigg\{\frac{a}{202}\bigg\}\equiv a\pmod{202}$.$$n+2n+\cdots kn\equiv 202\left(\frac{k}{2}\right)\pmod{202}\implies 101\mid \frac{k(k+1)n}{2}.$$Assume for the sake of contradiction that there exists $k>1$ when $n=101c$. We have $$\bigg\{\frac{c}{2}\bigg\}+\left\{\frac{2c}{2}\right\}+\cdots+\left\{\frac{kc}{2}\right\}=\frac{k}{2} \ .$$But $\bigg\{\frac{x}{2}\bigg\}=0$ or $\frac{1}{2}$ . By the size argument, we have $$\bigg\{\frac{c}{2}\bigg\}=\left\{\frac{2c}{2}\right\}=\cdots=\left\{\frac{kc}{2}\right\}=\frac{1}{2} \ .$$But since $k>1$, there must exist $\left\{\frac{2c}{2}\right\}=0$ in the sum which is absurd. Thus, if $101\mid n$ then $k=1$. Suppose that $101\nmid n$. Then $101\mid \frac{k(k+1)}{2}$ so $101\mid k$ or $101\mid k+1$. Since $k<202$, $k=100,101,201$. Thus, $k\in\{1,100,101,202\}$. It's easy to see that $(n,k)=(101,1),(2,100),(1,201)$ work. We claim that $(n,k)=(103,101)$ work. Note that for every $1\leq i\leq 101$,$$\text{Remainder of }103i\text{ divided by }202 \text{ are } \begin{cases} 2i+101 &\text{ for odd }1\leq i\leq 49. \\ 2i-101 &\text{ for odd }51\leq i\leq 101. \\2i &\text{ for even }2\leq i\leq 100. \end{cases}$$(The value of remainders are integers in the interval $[0,201]$.) We have \begin{align*}202\bigg(\left\{\frac{n}{202}\bigg\}+\left\{\frac{2n}{202}\right\}+\cdots+\left\{\frac{kn}{202}\right\}\right) &=\sum_{i=1}^{25} (2i-1+101)+\sum_{i=26}^{51}(2i-1-101)+\sum_{i=1}^{50}(2\cdot 2i)\\ &= 10201 \\ &= 202\left(\frac{k}{2}\right). \end{align*}Hence, $k=1,100,101,201$, as desired.

InternetPerson10 wrote: Just wanted to ask: what's the motivation for finding a value of $n$ that works for $k=101$? I spent the entire time convinced that nothing worked Python gives me loads of weird solutions, how do we find those? In contest, I found $n=51$ works by replacing $p$ with any 1 mod 4 prime, write $p=4k+1$, then $2k+1$ works where $k\in \{1,3\}$. So I guessed the construction $\frac{p+1}{2}$ works and it did The verification is left to the reader as an exercise Actually, $n=p+2$ is an easier construction

$n+2n+...+kn \equiv 101k$ (mod 202), then $\frac{k(k+1)}{2}n \equiv 101k$ (mode 202). Since $101$ is prime, it must satisfy at least $1$ out of $3$ below: $101 | k$, $101 | (k+1)$, $101 | n$. Then do some casework related with remainders.

Jalil_Huseynov wrote: Find all positive integers $k<202$ for which there exist a positive integers $n$ such that $$\bigg {\{}\frac{n}{202}\bigg {\}}+\bigg {\{}\frac{2n}{202}\bigg {\}}+\cdots +\bigg {\{}\frac{kn}{202}\bigg {\}}=\frac{k}{2}$$ We want $101|nk(k+1)/2$ so $k=100,101,201$ or $n=101$ for $n=101$ we have only $k=1$ for $k=100,201$ we can get $n=2 ,1$ and the dificult is for $k=101$ where we can get $n=103$ (I see it by CANBANKAN)

Let $t_i=(ni \pmod {202})$ for $i \in \{1,2,\ldots, k \}$. Note that $202\{\frac{ni}{202} \}=t_i$ for all $i$, and so the given condition rewrites as $$t_1+\ldots+t_k=101k$$ Moreover $0 \leq t_i<202$ for all $i$. Note that $t_i-it_1 \equiv ni-ni \equiv 0 \pmod {202}$, and so $$202 \mid (t_2-2t_1)+\ldots+(t_k-kt_1)=101k-t_1-2t_1-\ldots-kt_1=101k-\frac{k(k+1)}{2}t_1$$ Therefore, $101 \mid \frac{k(k+1)}{2}t_1$, and since $k,t_1<202$, we infer that $k=101$ or $k=100$ or $k=201$ or $t_1=101$ If $t_1=101$, then $n \equiv 101 \pmod {202}$ and we easily get $t_{2i} =0$ and $t_{2i+1}=101$ for all $i$, hence if $k$ is even we have $t_1+\ldots+t_k=101 \cdot \frac{k}{2} \neq 101k$, while if $k$ is odd we have $t_1+\ldots+t_k=101 \cdot \frac{k-1}{2}+101$, and this equals $101k$ only when $k=1$, and so $k=1$ with $n=101$ works. If $k=100$, we may take $n=2$, and so $t_i=2i$ for all $i$, hence $$t_1+\ldots+t_{100}=2(1+2+\ldots+100)=100 \cdot 101,$$ and so $k=100$ works. If $k=201$, we may take $n=1$, and so $t_i=i$ for all $i$, hence $t_1+\ldots+t_{201}=201 \cdot 101$, as desired. Lastly, if $k=101$ we may take $n=103$. In order to show this works, note that $t_{2i+1}=4i+103$ for all $i \leq 24$ and $t_{2i+1}=4i-99$ for all $25 \leq i \leq 50$, and $t_{2i}=4i$ for all $i$, hence $$t_1+\ldots+t_{101}=4(1+\ldots+50)+103 \cdot 25+4(0+1+\ldots+24)-26 \cdot 99+4(25+\ldots+50)=101^2,$$ as desired. To sum up, the only working $k$ are $1, 100,101$ and $201$.

We'll change the problem to find $k<202$ such that there exist a positive integers $n$ such that $\bigg {\{}\frac{n}{202}\bigg {\}}+\bigg {\{}\frac{2n}{202}\bigg {\}}+\cdots +\bigg {\{}\frac{kn}{202}\bigg {\}}-\frac{k}{2}$ is an integer, then we'll plug in the solution later. It's equivalent to: $\bigg {\{}\frac{n}{202}\bigg {\}}+2\bigg {\{}\frac{n}{202}\bigg {\}}+\cdots +k\bigg {\{}\frac{n}{202}\bigg {\}}-\frac{k}{2}= t$ $(t\in\mathbb Z)$ $\Rightarrow\frac{k(k+1)}{2}\bigg {\{}\frac{n}{202}\bigg {\}}-\frac{k}{2}=t$ $\Rightarrow k(k+1)\left(\frac{n}{202}-\left[\frac{n}{202}\right]\right)=2t+k$ $\Rightarrow k(k+1)\left(\frac{n}{202}\right)=2t+k+k(k+1)\left[\frac{n}{202}\right]$ It's easy to see that $k=1,100,101,201$ work now, I'll leave here for readers to try it out

Note that \[\frac{n}{202}+\frac{2n}{202}+\cdots+\frac{kn}{202}=\frac{nk(k+1)}{404}\equiv \frac{k}{2}\pmod{1}\] So $\frac{nk(k+1)}{404}\equiv \frac{202k}{404}\pmod 1$, so \[nk(k+1)\equiv 202k\pmod{404}\] This implies that $202\mid nk(k+1)$. Claim: Either $k=1$ or $202\mid k(k+1)$. Proof: Suppose $202\nmid k(k+1)$. Then also $101\nmid k(k+1)$, so $101\mid n$. Now we can WLOG $n\le 202$ because subtracting $n$ by $202$ doesn't do anything to the fractional parts. If $n=202$, then $\frac{k}{2}=0$, a contradiction. If $n=101$, then $\frac{k}{2}=\frac{1}{2}\cdot \lceil \frac{k}{2} \rceil\implies \lceil \frac{k}{2}\rceil =k$. This implies $\frac{k}{2}> k-1\implies k> 2k-2\implies k<2$, so $k=1$, as desired. $\blacksquare$ We have shown that everything except $\boxed{1,100,101,201}$ don't work. It remains to show that these values work. For $k=1$, set $n=101$. This works because $\{\frac{101}{202}\}=\frac{1}{2}$. For $k=100$, set $n=2$. This works because \[\left\{\frac{2}{202}\right\}+\left\{\frac{4}{202}\right\}+\cdots + \left\{\frac{200}{202}\right\}=\frac{10100}{202}=\frac{100}{2}\] For $k=201$, set $n=1$. This works because \[\sum_{i=1}^{201} \left\{\frac{i}{202}\right\}=\sum_{i=1}^{201} \frac{i}{202}=\frac{201\cdot 101}{202}=\frac{201}{2}\] For $k=101$, set $n=103$. Note that $2n\equiv 4\pmod{202}$ and $51n\equiv 1\pmod{202}$. Note that \[\sum_{i=1}^{50}\left\{\frac{2i}{202}\right\}=\frac{4+8+\cdots 200}{202}=\frac{5100}{202}\] Also, \[\sum_{i=0}^{25} \left\{\frac{2i+1}{202}\right\}=\frac{103+107+\cdots+ 199}{202}=\frac{3775}{202}\]and \[\sum_{i=25}^{50} \left\{\frac{2i+1}{202}\right\}=\frac{1+5+\cdots + 101}{202}=\frac{1326}{202}\] Now we add everything up. \[\sum_{i=0}^{101} \left\{\frac{i}{202}\right\}=\frac{5100+3775+1326}{202}=\frac{10201}{202}=\frac{101}{2},\]as desired. Thus, all of $1,100,101,201 $ work.

Very nice problem

The best approach to proving 101 works is by analogy with 9 seemingly

Solution : The solution of $k =1,100,101,201$. Case 1 : If $k=1$,then $n=101$ works. Case 2 : If $k=100$ ,then $n=2$ works. Case 3 : If $k=101$,then $n=103$ or $n=51$ . Case 4 : If $k=201$,then $n=1$ . Now for all other $k$, note $101\nmid k(k+1)$ but $\frac{k(k+1)n}{2\cdot 202} \equiv \frac k2 (\bmod\; 1)$ which means $n\equiv 0$ or $n\equiv 101 (\bmod\; 202)$. Observe that 202 does not divide $n$. If $n\equiv 101(\bmod\; 202)$ then carefully see that LHS is half the number of odd numbers in $[1,k]$. Therefore $k=1$.

All solutions are $k\in \left \{ 1,100,101,201\right \}$. Notice that $k(n(k+1)-202)=404a$ for $a\in \mathbb Z,$ and since LHS is a multiple of $101,$ it follows that $101$ divides at least one of numbers $k,n,(k+1).$ This trivially reduces the problem to case $k\in \left \{ 1,100,101,201\right \}$. By a straightforward check the following substitutions work: for $k=1$ take $n=101;$ for $k=100$ take $n=2;$ for $k=101$ take $n=103;$ for $k=201$ take $n=1.$

The only harder point in the problem is the case k=101

LHS $= \frac{nk(k+1)}{404} - \sum \frac{ni}{202} $ $\therefore \frac{nk(k+1)}{404} - \frac{k}{2}$ must be an integer, which implies that $k=101$ or $k \equiv -1 \pmod{101}$ or $101 \mid n$. Simple casework gives us the answer Edit: Top expression should have floor function in it. Too dumb to code it in latex.

attached sol

Let $a_i\equiv in\pmod{202}, 1\leq i\leq k$ Then $\sum_{i=1}^k \left\{\frac{in}{202}\right\}=\frac{1}{202}\sum_{i=1}^k a_i=\frac{k}{2}\implies \sum_{i=1}^k a_i=101k$ $\implies 101k\equiv\sum_{i=1}^k in\equiv \frac{1}{2}k(k+1)n\pmod{202}$ $\implies k(k+1)n\equiv202k\equiv0\pmod{202}\implies 202\vert k(k+1)n$ Case 1: $101\vert k(k+1)\implies k=100,101,201$ For $k=100$, $n=2$ works For $k=101$, $n=103$ works (why is left as an exercise for the reader) For $k=201$, $n=1$ works Case 2: $101\vert n\implies\sum_{i=1}^k \left\{\frac{in}{202}\right\}=\frac{\lceil\frac{k}{2}\rceil}{2}=\frac{k}{2}\implies k=1$ So the final answer is $k=1,100,101,201$

Suppose $k$ works. Note that $\left \{\frac{n+202}{202}\right \}=\left \{\frac{n}{202}\right \}$. Hence, we can assume $n<202$. For every $i\geq 1$, there must exists an integer $t_i\geq 0$ such that $$\left \{\frac{i\cdot n}{202}\right \}=\frac{n-202t_i}{202}=\frac{n}{202}-t_i.$$Let $t=t_1+t_2+\cdots+t_k$. We have $$\frac{n}{202} \cdot \frac{k(k+1)}2-t=\frac k2 \implies n=\frac{202(2t+k)}{k(k+1)} \qquad (1)$$ If $101\nmid k(k+1)$ then $k(k+1)\mid 2(2t+k)$ from $(1)$. Also, since $n<202$, we must have $2t+k<k(k+1)$. It follows that $k(k+1)=2(2t+k)$ (since there is a $c$ such that $c\cdot k(k+1)=2(2t+k)$ and $c\leq 2$ by the inequality). Hence $t=\frac{k(k-1)}4$ and $(1)$ implies that $n=101$. Note that $\{1/2\}+\{2/2\}+\dots+\{k/2\}=\begin{cases} k/4 & 2\mid k \\ (k+1)/4 & 2\nmid k \end{cases}$. It follows that $\frac{k+1}4=\frac k2$, i.e. $k=1$, this clearly works. Now, suppose $101\mid k(k+1)$ then $k\in \{101,100,201\}$. For $k=100$ take $n=4$, for $k=201$ take $n=3$ and for $n=101$ take $k=103$ and we can easily verify that they work.