= Solution
Let $s_k(t)$ be the shock separating the fans from $k-1/2$ and $k+1/2$. Their traces give
$$
\dot s_k=\frac12\left[\frac{s_k-k+1/2}{t}+\frac{s_k-k-1/2}{t}\right]=\frac{s_k-k}{t},\qquad s_k(1)=k+1/2.
$$
Thus $\boxed{s_k(t)=k+t/2}$. Between consecutive shocks, $u=(x-k+1/2)/t$. Writing $\{r\}=r-\lfloor r\rfloor$ for the <fractional part>, the period-one solution can be expressed, with right-hand shock traces, as
$$
\boxed{u(x,t)=\frac12+\frac{\{x-t/2\}-1/2}{t},\qquad t\geq1}.
$$
To put it in the requested fixed-support sum, take
$$
\boxed{v(x,t)=\begin{cases}\displaystyle\frac12+\frac{\{x-t/2\}-1/2}{t},&-1/2\leq x<1/2,\\0,&\text{otherwise},\end{cases}\qquad u(x,t)=\sum_{k\in\mathbb Z}v(x-k,t)}.
$$
The half-open cells partition the line, so exactly one summand contributes at each point. This is a representation of the periodic solution, not a superposition principle for arbitrary solutions of a nonlinear equation.
The left and right shock states are $1/2+1/(2t)$ and $1/2-1/(2t)$; both average to the shock speed $1/2$, and the strict inequalities certify causality. The <periodic square-wave relaxation for the inviscid Burgers equation> has the uniform estimate
$$
\boxed{\sup_x|u(x,t)-1/2|\leq\frac1{2t}\longrightarrow0}.
$$
The limit is the conserved mean $1/2$.
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