= Solution
For any $n\geq0$, take $f_n=T_{n+1}$, the <Chebyshev polynomial> of degree $n+1$. It has $n+2$ alternating extrema of magnitude one on $[-1,1]$. The <Chebyshev alternation theorem> shows that the zero polynomial is best from $\mathcal P_n$, with error one. Since it also lies in $\mathcal P_{n-1}$ when $n\geq1$,
$$
\boxed{E_{n-1}(T_{n+1})=E_n(T_{n+1})=1}.
$$
Now suppose $f^{(n)}(x)\gt0$ throughout $[-1,1]$ and, contrary to the claim, $E_{n-1}(f)=E_n(f)$. A best $p\in\mathcal P_{n-1}$ would then also be best in $\mathcal P_n$. Its error $e=f-p$ would have $n+2$ alternating extrema by the alternation theorem, hence at least $n+1$ distinct zeros. Applying the <Rolle theorem> $n$ times gives a zero of
$$
e^{(n)}=f^{(n)}-p^{(n)}=f^{(n)},
$$
contradicting positivity. Therefore
$$
\boxed{E_{n-1}(f)\gt E_n(f)}.
$$
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