= Solution
The original PDF has indices $3^k$; the exponent is lost in the TeX transcription. This lacunary indexing is essential to the <positive lacunary Chebyshev series> argument below.
Let $K$ be the least nonnegative integer with $3^K>n$, so $K=0$ for $n=0$. Define
$$
p_n(x)=\sum_{k=0}^{K-1}a_kT_{3^k}(x),\qquad
A_K=\sum_{k=K}^{\infty}a_k.
$$
For $K=0$ the sum is empty and means the zero <polynomial>. Each included <Chebyshev polynomial> has degree at most $n$. Also $|T_j(x)|\le1$ on the interval, so summability of the positive coefficients gives <uniform convergence> by the <Weierstrass M-test>, and
$$
\|f_0-p_n\|_\infty\le A_K.
$$
Choose $L=3^K$ and the $L+1$ points $x_j=\cos(j\pi/L)$, $0\le j\le L$. For every omitted index $k\ge K$, the integer $3^{k-K}$ is odd. Consequently,
$$
T_{3^k}(x_j)=\cos(3^{k-K}j\pi)=(-1)^j.
$$
Every term of the tail has the same sign at a given point, and therefore
$$
f_0(x_j)-p_n(x_j)=(-1)^jA_K.
$$
This shows both that the error norm is exactly $A_K$ and that it alternates at $L+1\ge n+2$ distinct points. The points are in decreasing order; reversing their order still gives alternation. The <Chebyshev alternation theorem> proves that this partial sum is the unique <best uniform approximation>. Hence
$$
\boxed{p_n=\sum_{3^k\le n}a_kT_{3^k},\qquad
E_n(f_0)=\sum_{3^k>n}a_k}.
$$
In particular, $p_0=0$ and $E_0(f_0)=\sum_{k=0}^\infty a_k$. The equal signs of all tail terms at the same extrema are the reason positivity and the odd integer frequency ratios are useful.
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