Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Consider each of the $ m^n$ integers $ \sum e_ia_i$ with each of the $ e_i$ varying in $ \{0,1,\ldots,m-1\}$. If any two have the same residue $ \pmod{m^n}$ then we take their difference and we are done. Otherwise, they form a complete residue class $ \pmod{m^n}$ (1) (with the all-zeroes case corresponding to $ 0\pmod{m^n}$). We show this is a contradiction. Let $ b_i$ be the number of these numbers $ i \pmod{m^n}$ Take $ \epsilon=e^{2i\pi/m^n}$ and consider $ \sum b_i\epsilon^i=\sum_{e_i\in\{0,1,\ldots,m-1\}} \epsilon^{e_1a_1+\ldots+e_{m^n}a_{m^n}}=\prod \frac{\epsilon^{ma_i}-1}{\epsilon^{a_i}-1}=0$ by assumption (1) (note that the denominators cannot be zero by the problem condition). But then $ \epsilon^{ma_i}=1$ for some $ i$, and hence $ m^{n-1}|a_i$; contradiction.

interpretation ?

Similar problem was appeared in Romanian TST https://artofproblemsolving.com/community/c6h53557p335760

I really enjoyed this problem! For simplicity, let $[a,b]$ be the set of integers $a\leq k\leq b.$ For each vector $v=(x_1,x_2,...,x_n)\in\mathbb{R}^n$ define $f(v):=x_1\cdot a_1+x_2\cdot a_2+...+x_n\cdot a_n.$ In what follows, we will only work in $\mathbb{F}_{m^n}.$ Begin by assuming that no non-zero vector $v\in[-(m-1),m-1]^n$ satisfies $f(v)=0.$ Let $u,v\in [0,m-1]^n$ be two distinct, non-zero vectors. If $f(v)=f(u)$ then, by the linearity of $f,$ we get that for $w=u-v\in [-(m-1),m-1]^n,$ $f(w)=0.$ This is a contradiction to our assumption. Therefore, $\{0\}\cup\{f(v): v\neq 0, \ v\in [0,m-1]^n\}$ is a complete residue set mod $m^n.$ However, since $f(0)=0$ we can more simply say that $\{f(v): v\in[0,m-1]^n\}$ is a complete residue set modulo $m^n.$ Thus \[\prod_{i=1}^n\frac{X^{a_i\cdot m}-1}{X^{a_i}-1}=\prod_{i=1}^n\big(X^{a_i\cdot 0}+X^{a_i\cdot 1}+...+X^{a_i\cdot (m-1)}\big)=\sum_{v\in [0,m-1]^n}X^{f(v)}=\sum_{i=0}^{m^n-1}X^i=\frac{X^{m^n}-1}{X-1}.\]By letting $X$ be the $m^n$th root of unity and using the fact that $m^{n-1}\nmid a_i,$ we get a contradiction.

Solved with a hint It suffices to show that there exist two choices of $(b_1,\ldots,b_n)$ with $0 \leq b_i<m$ for all $i$ such that $b_1a_1+\cdots+b_na_n$ collide modulo $m$, whence we can take $e_i$ to be the termwise difference between these two sequences. If no collisions occur, because there are $m^n$ choices of a sequence, $b_1a_1+\cdots+b_na_n$ must cover every residue modulo $m^n$ exactly once. Thus if we let $z=e^{2\pi i/m^n}$ be a primitive $m^n$-th root of unity, we should have $$0=\sum_{(b_1,\ldots,b_n)} z^{b_1a_1+\cdots+b_na_n}=\prod_{i=1}^n \sum_{j=0}^{m-1} z^{ja_i}=\prod_{i=1} \frac{z^{ma_i}-1}{z^{a_i}-1}.$$Thus some $i$ must exist with $m^n \mid ma_i \implies m^{n-1} \mid a_i$, which is illegal. Thus a collision must occur, so we are done. $\blacksquare$

interesting

Suppose there were two distinct sequences $x_1,x_2,\dots,x_n$ and $y_1,y_2,\dots,y_n$ such that $0\le x_i,y_i\le m-1$ for all $i$ and \[x_1a_1+x_2a_2+\dots+x_na_n\equiv y_1a_1+y_2a_2+\dots+y_na_n\pmod {m^n}\]then we are done by setting $e_i=x_i-y_i$ for all $i$. Suppose for all sequences $x_1,x_2,\dots,x_n$ such that $0\le x_i\le m-1$, the remainder when dividing $m^{n}$ is distinct. Then, it must reach all possible remainders because there are $m^n$ total sequences. Now, let $z$ be a primitive $m^n$th root of unity. We have that the sum over all sequences $x$ of $z^{x_1a_1+x_2a_2+\dots+x_na_n}$ is equal to $0$. Let $z^{a_1}i=z_i$ for all $z$. Note that $z_i$ is not a $m$th root of unity for any $i$. Then, we are simply adding up \[\prod_{i=1}^{n}{z_i^{x_i}}\]and it is easy to see that this factors into \[\prod_{i=1}^{n}(1+z_i+z_i^2+\dots+z_i^{m-1})=\prod_{i=1}^{n}\left(\frac{z_i^{m}-1}{z_i-1}\right)\]But this is definitely not zero.