Denote by $\delta_n$ the partition $(n-1,n-2,\dots,1)$ of $\binom{n}2$, and the associated schur function is
$$s_{\delta_n}(x_1,\dots,x_n)=\prod_{i<j}(x_i+x_j).$$
Consider now the ratio of determinants
$$\begin{vmatrix}
x_1^{2r-1} & x_2^{2r-1} &\cdots & x_n^{2r-1} \\
x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\
\vdots &\vdots & \ddots & \vdots\\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
1 & 1& \cdots &1
\end{vmatrix} \cdot \begin{vmatrix}
x_1^{2n-2} & x_2^{2n-2} &\cdots & x_n^{2n-2} \\
x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\
\vdots &\vdots & \ddots & \vdots\\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
1 & 1& \cdots &1
\end{vmatrix}^{-1}$$
This evaluates to $\sum_{k=1}^nx_k^{2r-1}\prod_{\ell\neq k}^{1,n}\frac1{x_{\ell}^2-x_k^2}$. Proof: Expand the first determinant by the first row, and use the fact that the second is just a Vandermonde that evaluates to $\prod_{i<j} (x_j^2-x_i^2)$.

Therefore your polynomial $P_r(x_1,x_2,\dots,x_n)$ is equal to
$$(-1)^r \begin{vmatrix}
x_1^{2r} & x_2^{2r} &\cdots & x_n^{2r} \\
x_1^{2n-3} & x_2^{2n-3} &\cdots & x_n^{2n-3} \\
\vdots &\vdots & \ddots & \vdots\\
x_1^3 & x_2^3 & \cdots & x_n^3 \\
x_1 & x_2& \cdots &x_n
\end{vmatrix} \cdot \begin{vmatrix}
x_1^{2n-2} & x_2^{2n-2} &\cdots & x_n^{2n-2} \\
x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\
\vdots &\vdots & \ddots & \vdots\\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
1 & 1& \cdots &1
\end{vmatrix}^{-1}\cdot s_{\delta_n}(x_1,x_2,\dots,x_n)$$
When $0\le r\le n$ we can use the definition of Schur polynomials as ratios of determinants to simplify this expression as
$$s_{\lambda_{r,n}}(x_1,x_2,\dots,x_n)$$
Where the partition $\lambda_{r,n}$ is given by $(n-2,n-3,\dots,r+1,r,r,r,r-1,\dots,2,1)$. Therefore, not only is $P_{r}(x_1,x_2,\dots,x_n)$ symmetric with positive integer coefficients, but we also get a combinatorial interpretation of these coefficients as counting semistandard tableaux of shape $\lambda_{r,n}$.