= Solution
Suppose for contradiction that some $q\in\mathcal P_n$ satisfies
$$
\|f-q\|_\infty\lt a_*:=\min_i a_i.
$$
At $t_i$,
$$
p(t_i)-q(t_i)=[f(t_i)-q(t_i)]-(-1)^ia_i.
$$
Because $|f(t_i)-q(t_i)|\lt a_i$, the values $p(t_i)-q(t_i)$ have strictly alternating signs. The <intermediate value theorem> therefore gives at least one root of $p-q$ in each of the $n+1$ intervals $(t_i,t_{i+1})$. A nonzero polynomial of degree at most $n$ cannot have $n+1$ distinct roots. If $p-q$ were identically zero, its error at $t_i$ would be $a_i\geq a_*$, also a contradiction. Hence
$$
\boxed{E_n(f)\geq\min_{1\leq i\leq n+2}a_i}.
$$
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