Answer is max{$\frac{1}{a},\frac{n}{n-1+a}$} If a<1,take $x_{1}=x_{2}=...=x_{n-1}=k,x_{n}=\frac{1}{k^{n-1}} :k->0$ $=> \sum_{i=1}^{n}\frac{1}{a+S-x_{i}}=\frac{1}{a+(n-1)k}+\frac{n-1}{a+(n-2)k+\frac{1}{k^{n-1}}}->\frac{1}{a}$ so we will prove that $ \sum_{i=1}^{n}\frac{1}{a+S-x_{i}}\leq\frac{1}{a}$ Denote $a_{1},a_{2},...,a_{n}$ for which $a_{i}^{n-1}=x_{i}$ and $a_{1}+a_{2}+...+a_{n}=T$ $S-x_{i}\geq AM-GM\geq (T-a_{i}).\frac{1}{a_{i}} ...(1)$ $a.a_{i}+a_{i}(S-x_{i})\geq a.a_{i}+T-a_{i}\geq a.a_{i}+a.(T-a_{i})=a.(a_{1}+a_{2}+...+a_{n})$ $ => \frac{1}{a+S-x_{i}}\leq \frac{a_{i}}{a.(a_{1}+a_{2}+...+a_{n})} => \sum_{i=1}^{n}\frac{1}{a+S-x_{i}}\leq \frac{1}{a} $ If $a\geq 1$ take $x_{1}=x_{2}=...=x_{n}=1 =>M=\frac{n}{n-1+a}$ $(a_{i})(a+S-a_{i})\geq (1)\geq a.a_{i}+T-a_{i} =>\frac{1}{a+S-a_{i}}\leq \frac{a_{i}}{a.a_{i}+T-a_{i}} =>$ $ \sum_{i=1}^{n}\frac{1}{a+S-x_{i}}\leq \sum_{i=1}^{n}\frac{a_{i}}{a.a_{i}+T-a_{i}}\leq \frac{n}{n-1+a} <=>$ $ \sum_{i=1}^{n}\frac{T-a_{i}}{a.a_{i}+T-a_{i}}\geq n-\frac{a.n}{n-1+a}=\frac{n.(n-1)}{n-1+a} $ $ \sum_{i=1}^{n}\frac{T-a_{i}}{a.a_{i}+T-a_{i}}= \sum_{i=1}^{n}\frac{(T-a_{i})^{2}}{(T-a_{i})(a.a_{i}+T-a_{i})}\geq couchy\geq \frac{(n-1)^{2}.(a_{1}+a_{2}+...+a_{n})^{2}}{ \sum_{i=1}^{n} (T-a_{i})^{2}+a.a_{i}.(T-a_{i})}$ $\geq \frac{n.(n-1)}{n-1+a} <=> {(n-1).(a_{1}+a_{2}+...+a_{n})^{2}.(n-1+a)\geq \sum_{i=1}^{n} (T-a_{i})^{2}+a.a_{i}.(T-a_{i}) }$ and this is just AM-GM (use that $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\geq \frac{2}{n-1}. \sum_{1\leq i<j\leq n}^{n}a_{i}.a_{j} ) $