For symmetric data
$(x_i,y_i), i=-n,-n+1,..., n-1, n$
such that
$x_{-i}=-x_i$ and $y_{-i}=-y_i, i=0,1,...n$
what is the required degree for an interpolating polynomial $p$? Since there are $2n+1$ data points (one for each $\pm n$ plus one for when $i=0$) would the degree be at most $2n$? Or would only the points when $i\ge0$, be considered for a degree at most $n$?
Also, to show that the polynomial is odd (which is information given) so $-p(x)=p(-x)$, is it sufficient to show that since $p(x_{-i})=p(-x_i)$, then $p(-x_i)=-p(x_i)$?
Well, that is frequently a confusing issue. Let me try and organize it in this way.
In general, $2n+1$ points will be interpolated by a polynomial of degree at most $2n$.
When, like in this case, the $2n+1$ points $$ \left( {x_{\,i} ,y_{\,i} } \right)\quad \left| {\;\left\{ \matrix{ i = - n, \ldots ,0, \ldots ,n \hfill \cr x_{\,i} = - x_{\,i} \hfill \cr y_{\,i} = - y_{\,i} \hfill \cr} \right.} \right.\quad \to \quad \left( {x_{\,0} ,y_{\,0} } \right) = \left( {0,0} \right) $$ constitute an anti-symmentric pattern, then they will be interpolated by a odd polynomial $$ y(x)\quad \left| {\;y( - x) = - y(x)} \right.\quad \to \quad y(x) = \sum\limits_{1\, \le \,k\, \le \;n} {c_{\;2k - 1} x^{\,2k - 1} } $$
of degree (at most) $2n-1$ , thus having only $n$ undetermined coefficients, which you will set by imposing to pass through $n$ points.
These can be obviously the points with $i = 1, \ldots ,n$ (but not necessarily, they could also be chosen as $ i = \pm k\quad \left| {\;k = 1, \ldots ,n} \right.$ ). The $n+1$ even coefficient $c_{\;0} ,c_{\;2} ,\, \ldots ,c_{\;2n} $ are null due to the anti-simmetry.
If the points where instead symmetric then the polynomial would be of degree (<=) $2n$, with $n+1$ undetermined coefficients. The additional indetermination ($c_{\;0} $) being due to that in this case ${y_{\,0} }$ is not fixed.
In conclusion, you can always apply the general method of interpolation. But when there is a symmetry, you can profitably reduce the interpolation and consider only the "independent" points.