Most surely, $$max=\frac1{1+n}+\sum_{i=2}^n \frac i{i+(i-1)},$$ but my proof below is still somewhat incomplete...

If the $x_i$ are a permutation $\pi$ of $\{1,2,...,n\}$, let $y_i:=\pi^{-1}(i)$ Then we have $$\sum_{i=1}^n \frac{i}{i+x_i}=n-\sum_{i=1}^n \frac{x_i}{i+x_i}=n-\sum_{j=1}^n \frac{j}{y_j+j},$$ so the sums come in pairs, and if $\pi$ yields a maximum, then $\pi^{-1}$ will yield a minimum.

I'd like to show that the maximum is attained for the permutation ${\pi:=(n,1,2,...,n-1)}$.

Now, for $i>1$ and $k<n$, $$\frac1{1+n}+\frac i{i+k}>\frac 1{1+k}+\frac i{i+n}$$because $$LHS-RHS=\frac{(i - 1) (n-k) (k n - i)}{(1+n) (i + k)(1+k) (i + n)}>0.$$ So we must have $x_1=n$ for a maximal permutation.

Generally, $$\frac j{j+a}+\frac k{k+b}-\left(\frac j{j+b}+\frac k{k+a}\right) =\frac{(b-a)(k-j)(jk-ab)}{(j+a) (k+b) (j+b)(k+a)}.$$ Thus if for $j\ge2$ we put $x_j=j-1$, the above difference (with putting $a=x_j, b=x_k$) is always positive. This proves that $\pi$ does strictly better than with any involution applied to it, which gives a strong evidence in favor of $\pi$.
But the problem remains that we cannot conclude from there for any combination of involutions because of the factor $(jk-ab)$ which might become negative at some point.

EDIT after js21's answer: Yes, it looks indeed like the maximum is attained for the permutation $$\color{red}{[n,n-1,...,n-k+1,1,2,...,n-k]}$$ with $k\approx \left\lfloor\dfrac n{6.3031}\right\rfloor$, thus linear in $n$. And this is very precise: for $n<500,000$, the optimal $k$ is within $\pm1$ of this, even for small $n$!

Most surely, (see bottom) Initially I had thought $$max=\frac1{1+n}+\sum_{i=2}^n \frac i{i+(i-1)},$$ but my proof below is still somewhat incomplete...

If the $x_i$ are a permutation $\pi$ of $\{1,2,...,n\}$, let $y_i:=\pi^{-1}(i)$ Then we have $$\sum_{i=1}^n \frac{i}{i+x_i}=n-\sum_{i=1}^n \frac{x_i}{i+x_i}=n-\sum_{j=1}^n \frac{j}{y_j+j},$$ so the sums come in pairs, and if $\pi$ yields a maximum, then $\pi^{-1}$ will yield a minimum.

I'd like to show that the maximum is attained for the permutation ${\pi:=(n,1,2,...,n-1)}$.

Now, for $i>1$ and $k<n$, $$\frac1{1+n}+\frac i{i+k}>\frac 1{1+k}+\frac i{i+n}$$because $$LHS-RHS=\frac{(i - 1) (n-k) (k n - i)}{(1+n) (i + k)(1+k) (i + n)}>0.$$ So we must have $x_1=n$ for a maximal permutation.

Generally, $$\frac j{j+a}+\frac k{k+b}-\left(\frac j{j+b}+\frac k{k+a}\right) =\frac{(b-a)(k-j)(jk-ab)}{(j+a) (k+b) (j+b)(k+a)}.$$ Thus if for $j\ge2$ we put $x_j=j-1$, the above difference (with putting $a=x_j, b=x_k$) is always positive. This proves that $\pi$ does strictly better than with any involution applied to it, which gives a strong evidence in favor of $\pi$.
But the problem remains that we cannot conclude from there for any combination of involutions because of the factor $(jk-ab)$ which might become negative at some point.

EDIT after js21's answer: That shot was too quick. It looks in fact like the maximum is attained for the permutation $$\color{red}{\pi_k=[n,n-1,...,n-k+1,1,2,...,n-k]}$$ with $k\approx \dfrac n{6.3032}$, thus linear in $n$. This is very precise: for $n<1,000,000$, the optimal $k$ always seems to be within a range $\pm1$ of this fraction, even for small $n$.

Most surely, $$max=\frac1{1+n}+\sum_{i=2}^n \frac i{i+(i-1)},$$ but my proof below is still somewhat incomplete...

If the $x_i$ are a permutation $\pi$ of $\{1,2,...,n\}$, let $y_i:=\pi^{-1}(i)$ Then we have $$\sum_{i=1}^n \frac{i}{i+x_i}=n-\sum_{i=1}^n \frac{x_i}{i+x_i}=n-\sum_{j=1}^n \frac{j}{y_j+j},$$ so the sums come in pairs, and if $\pi$ yields a maximum, then $\pi^{-1}$ will yield a minimum.

I'd like to show that the maximum is attained for the permutation $\color{red}{\pi:=(n,1,2,...,n-1)}$.

Now, for $i>1$ and $k<n$, $$\frac1{1+n}+\frac i{i+k}>\frac 1{1+k}+\frac i{i+n}$$because $$LHS-RHS=\frac{(i - 1) (n-k) (k n - i)}{(1+n) (i + k)(1+k) (i + n)}>0.$$ So we must have $x_1=n$ for a maximal permutation.

Generally, $$\frac j{j+a}+\frac k{k+b}-\left(\frac j{j+b}+\frac k{k+a}\right) =\frac{(b-a)(k-j)(jk-ab)}{(j+a) (k+b) (j+b)(k+a)}.$$ Thus if for $j\ge2$ we put $x_j=j-1$, the above difference (with putting $a=x_j, b=x_k$) is always positive. This proves that $\pi$ does strictly better than with any involution applied to it, which gives a strong evidence in favor of $\pi$.
But the problem remains that we cannot conclude from there for any combination of involutions because of the factor $(jk-ab)$ which might become negative at some point.

Most surely, $$max=\frac1{1+n}+\sum_{i=2}^n \frac i{i+(i-1)},$$ but my proof below is still somewhat incomplete...

If the $x_i$ are a permutation $\pi$ of $\{1,2,...,n\}$, let $y_i:=\pi^{-1}(i)$ Then we have $$\sum_{i=1}^n \frac{i}{i+x_i}=n-\sum_{i=1}^n \frac{x_i}{i+x_i}=n-\sum_{j=1}^n \frac{j}{y_j+j},$$ so the sums come in pairs, and if $\pi$ yields a maximum, then $\pi^{-1}$ will yield a minimum.

I'd like to show that the maximum is attained for the permutation ${\pi:=(n,1,2,...,n-1)}$.

Now, for $i>1$ and $k<n$, $$\frac1{1+n}+\frac i{i+k}>\frac 1{1+k}+\frac i{i+n}$$because $$LHS-RHS=\frac{(i - 1) (n-k) (k n - i)}{(1+n) (i + k)(1+k) (i + n)}>0.$$ So we must have $x_1=n$ for a maximal permutation.

Generally, $$\frac j{j+a}+\frac k{k+b}-\left(\frac j{j+b}+\frac k{k+a}\right) =\frac{(b-a)(k-j)(jk-ab)}{(j+a) (k+b) (j+b)(k+a)}.$$ Thus if for $j\ge2$ we put $x_j=j-1$, the above difference (with putting $a=x_j, b=x_k$) is always positive. This proves that $\pi$ does strictly better than with any involution applied to it, which gives a strong evidence in favor of $\pi$.
But the problem remains that we cannot conclude from there for any combination of involutions because of the factor $(jk-ab)$ which might become negative at some point.

EDIT after js21's answer: Yes, it looks indeed like the maximum is attained for the permutation $$\color{red}{[n,n-1,...,n-k+1,1,2,...,n-k]}$$ with $k\approx \left\lfloor\dfrac n{6.3031}\right\rfloor$, thus linear in $n$. And this is very precise: for $n<500,000$, the optimal $k$ is within $\pm1$ of this, even for small $n$!