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For $2k$ real numbers $a_1, a_2, ..., a_k$, $b_1, b_2, ..., b_k$ define a sequence of numbers $X_n$ by
\[X_n = \sum_{i=1}^k [a_in + b_i] \quad (n=1,2,...).\] If the sequence $X_N$ forms an arithmetic progression, show that $\textstyle\sum_{i=1}^k a_i$ must be an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.

(This is a solution I saw later in the math camp.)
Let $\displaystyle\sum_{i=1}^{k}a_i=A,\displaystyle\sum_{i=1}^{k}b_i=B$. Also let $X_i=X_1+(i-1)d$. Since $X_1,X_2$ are integers, $d=X_2-X_1$ is also an integer.
Then, $\displaystyle\sum_{i=1}^{k}(a_in+b_i+1)>\displaystyle\sum_{i=1}^{k}\lfloor a_in+b_i\rfloor>\displaystyle\sum_{i=1}^{k}(a_in+b_i-1)$
$\Rightarrow nA+B+k>X_n>nA+B-k...........................(i)$
Putting $n=1$ in $(i)$ we deduce that $A+B+k>X_1>A+B-k$
$\Rightarrow -A-B+k<-X_1<-A-B-k...........................(ii)$
Adding (i) and (ii), $(n-1)A+2k>X_n-X_1>(n-1)A-2k$
$\Rightarrow 2k>(n-1)(d-A)>-2k$.
Now if $d-A\neq 0$ then taking $n\to \infty$ will make $(n-1)(d-A)>2k$ or $(n-1)(d-A)<-2k$ so we must have
$d-A=0\Rightarrow A=d\in \mathbb{Z}$.

Phlembac Adib Hasan wrote:For $2k$ real numbers $a_1, a_2, ..., a_k$, $b_1, b_2, ..., b_k$ define a sequence of numbers $X_n$ by
\[X_n = \sum_{i=1}^k [a_in + b_i] \quad (n=1,2,...).\] If the sequence $X_N$ forms an arithmetic progression, show that $\textstyle\sum_{i=1}^k a_i$ must be an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.

As always, let $\displaystyle\sum_{i=1}^{k}a_i=A,\displaystyle\sum_{i=1}^{k}b_i=B$, and let $X_n = \sum_{i=1}^k [a_in + b_i] = Sn+T$ (as it is an arithmetic series).
So, we have $\sum (a_in+b_i) \geq \sum [a_in + b_i] \geq \sum (a_in + b_i -1) $
$\Rightarrow An +B \geq Sn+T \geq An+B - k$
$\Rightarrow A+ \frac B n \geq S +\frac Tn \geq A + \frac Bn - \frac kn$

Taking $n \rightarrow \infty$ implies $S = A$, and as $S$ is an integer, so is $A$.