Exist $a,b$ such that $-1<a<0<b<1$ and satisfy $|P(a)| \geq 1$, $P(b) \geq 1$

Let's number the roots so $x_1=z_1,..x_k=z_k \ge 0$ and $y_1=-x_{k+1},...y_{2n-k}=-x_{2n} \ge 0$, where $0 \le k \le 2n$ (if $k=0$ or $k=2n$ the corresponding set is empty).

Assume there is $-1<a<0, 0<b<1$ st $|P(a)| \ge 1, |P(b)| \ge 1$ (which clearly implies that $0<k<2n$ since the roots cannot be all negative or all non-negative by the assumption above)

$|P(a)P(b)|=\Pi_{1 \le r \le k, 1 \le q \le 2n-k}(|a|+z_r)|(b-z_r)|(|a|-y_q)|(b+y_q)$, so using the mean inequality we get:

$4n|P(a)P(b)|^{4n} \le \sum (|a|+z_r)+(b+y_q)+|b-z_r|+||a|-y_q|=l|a|+mb+\sum c_rz_r+\sum d_qy_q=S$,

where $l+m+\sum c_r+\sum d_q=4n$ since when we resolve the absolute values the sum of the coefficients is zero, while the sum of the coefficients in the sums is just $2k+2(2n-k)=4n$ as $r$ takes $k$ values and $q$ takes $2n-k$ values (note that by assumption $P(a)P(b) \ne 0$ so $a,b$ are not roots, hence the absolute values are unambiguously resolved)

Note that $c_r,d_q$ are either $0$ or $2$ while $l \ge 2k-2n, m \ge 2n-2k$ so one of $l,m >0$ unless $k=n$

Now if $l, m \ge 0$ then since $0 \le z_r, y_q \le 1$ and $0<|a|,b<1$, we get $S < l+m+\sum c_r +\sum d_q < 4n$ unless $l=m=0$ (hence $k=n$ as above) and then again $S<4n$ unless $z_r=y_q=1$ which implies $P(x)=(1-x^2)^n$ for which $|P(a)|, |P(b)|<1$ contradiction.

Otherwise $S<4n$ in this case, so $|P(a)P(b)|<1$ Contradiction again!

Now assume one of $l,m <0$ and wlog we can assume $m<0$ hence $k>n$ so there are at least $2n-k+1$'s $z_r$'s where $|b-z_r|=z_r-b$, hence numbering $2n-k$ of those as $w_1,..w_{2n-k}$ we note that $P(b)=Q(b)\Pi_{q=1,2n-k}(b-w_q)(b+y_q)$ with $|Q(b)| <1$ as it consists only of factors $b-z_r$ which are in $(-1,1)$ and there is at least one such since $k>n$

But now we have that $0<b<w_q$ for all $q=1,..2n-k$ and the quadratic $A(x)=(w_q-x)(x+y_q)$ assumes its absolute value maximum on $[0,w-q]$ either at its ends or at $\frac{y_q-w_q}{2}$ (if it's in the interval which means $w_q \le y_q \le 3w_q$) and $|A(0)| \le 1, A(w_q)=0, |A(\frac{y_q-w_q}{2})| =\frac{3w_q-y_q}{2} \frac{3y_q-w_q}{2} \le ((w_q+y_q)/2)^2 \le 1$ by the mean inequality, so in any case each product $|(w_q-b)(b+y_q)| \le 1$ and combined with $|Q(b)|<1$ gives $|P(b)|<1$ and we get a contradiction in this case again!