Straightforward exercise for anyone used to linear algebra Number the persons $1$ through $n$ and consider the $n\times n$ $(0,1)$-matrix given by $A=(a_{ij})_{1\le i,j\le n}$ with $a_{ij}=1$ if $j$ follows $i$ and $0$ otherwise. The main observation is that if in midnight the column vector given by the user's rating is $x$, the next midnight it will be $Ax.$ Let $e_i\in\mathcal{M}_{n,1}$ be the column vector with $1$ in position $i$ and $0$ in all other positions. The hacker can add one $e_i$ to $x$ at midnight, or can do nothing. Let $a_1,a_2,...,a_n$ be the initial user's ratings reduced modulo $m$ and $b_i=m-a_i$ for each $i$. By PHP, there are distinct numbers $p>c_1>c_2>...>c_{b_1+...+b_n}$ such that $A^p\equiv A^{c_j}(\text{mod}~m)$ for any $j$. Then $$(a_i\cdot A^p+A^{c_{b_1+...+b_{i-1}+1}}+....+A^{c_{b_1+....+b_i}})e_i\equiv 0(\text{mod}~m)$$for each $i=\overline{1,n}.$ Thus the hacker can add $e_i$ in midnights $p-c_{b_1+...+b_{i-1}+1},...,p-c_{b_1+....+b_i}$ for each $i$, and by the above it can easily be seen to be enough, since the resulting vector would be $$A^px+\sum_{i=1}^{n}(A^{c_{b_1+...+b_{i-1}+1}}+....+A^{c_{b_1+....+b_i}})e_i=\sum_{i=1}^{n} (a_i\cdot A^p+A^{c_{b_1+...+b_{i-1}+1}}+....+A^{c_{b_1+....+b_i}})e_i\equiv 0(\text{mod}~m).$$

Can someone explain me more about what above said?