28. Note that w.l.o.g., we can assume that $p$ and $q$ are coprime. Indeed, otherwise it suffices to consider the problem in which all $x_{i}$'s and $p, q$ are divided by $\operatorname{gcd}(p, q)$. Let $k, l$ be the number of indices $i$ with $x_{i+1}-x_{i}=p$ and the number of those $i$ with $x_{i+1}-x_{i}=-q$ (0 \leq i < n). We have $k+l=n$, $kp-lq=0$, and thus $k=qt$, $l=pt$, and $n=(p+q)t$. Consider the sequence $y_{i}=x_{i+p+q}-x_{i}$, $i=0, \ldots, n-p-q$. We claim that at least one of the $y_{i}$'s equals zero. We begin by noting that each $y_{i}$ is of the form $up-vq$, where $u+v=p+q$; therefore $y_{i}=(u+v)p-v(p+q)=(p-v)(p+q)$ is always divisible by $p+q$. Moreover, $y_{i+1}-y_{i}=\left(x_{i+p+q+1}-x_{i+p+q}\right)-\left(x_{i+1}-x_{i}\right)$ is 0 or $\pm(p+q)$. We conclude that if no $y_{i}$ is 0 then all $y_{i}$'s are of the same sign. But this is in contradiction with the relation $y_{0}+y_{p+q}+\cdots+y_{n-p-q}=x_{n}-x_{0}=0$. Consequently some $y_{i}$ is zero, as claimed.

Second solution. As before we assume $(p, q)=1$. Let us define a sequence of points $A_{i}\left(y_{i}, z_{i}\right)(i=0,1, \ldots, n)$ in $\mathbb{N}_{0}^{2}$ inductively as follows. Set $A_{0}=(0,0)$ and define $\left(y_{i+1}, z_{i+1}\right)$ as $\left(y_{i}, z_{i}+1\right)$ if $x_{i+1}=x_{i}+p$ and $\left(y_{i}+1, z_{i}\right)$ otherwise. The points $A_{i}$ form a trajectory $L$ in $\mathbb{N}_{0}^{2}$ continuously moving upwards and rightwards by steps of length 1. Clearly, $x_{i}=pz_{i}-qy_{i}$ for all $i$. Since $x_{n}=0$, it follows that $\left(z_{n}, y_{n}\right)=(kq, kp)$, $k \in \mathbb{N}$. Since $y_{n}+z_{n}=n>p+q$, it follows that $k>1$. We observe that $x_{i}=x_{j}$ if and only if $A_{i}A_{j} \| A_{0}A_{n}$. We shall show that such $i, j$ with $i<j$ and $(i, j) \neq (0, n)$ must exist. If $L$ meets $A_{0}A_{n}$ in an interior point, then our statement trivially holds. From now on we assume the opposite. Let $P_{ij}$ be the rectangle with sides parallel to the coordinate axes and with vertices at $(ip, jq)$ and $((i+1)p, (j+1)q)$. Let $L_{ij}$ be the part of the trajectory $L$ lying inside $P_{ij}$. We may assume w.l.o.g. that the endpoints of $L_{00}$ lie on the vertical sides of $P_{00}$. Then there obviously exists $d \in \{1, \ldots, k-1\}$ such that the endpoints of $L_{dd}$ lie on the horizontal sides of $P_{dd}$. Consider the translate $L_{dd}^{\prime}$ of $L_{dd}$ for the vector $-d(p, q)$. The endpoints of $L_{dd}^{\prime}$ lie on the vertical sides of $P_{00}$. Hence $L_{00}$ and $L_{dd}^{\prime}$ have some point $X \neq A_{0}$ in common. The translate $Y$ of point $X$ for the vector $d(p, q)$ belongs to $L$ and satisfies $XY \| A_{0}A_{n}$.