Let $A$ be Hermitian. Assume that the eigenvalues of $A$ are increasing ordered and that $a_{ii} = \lambda_n$. We have that $$\sum_{j=1}^{m}a_{ii} \geq \sum_{j=1}^{m}\lambda_j$$ and $$\sum_{j=1}^{m}a_{ii} \leq \sum_{j=n-m+1}^{n}\lambda_j$$ How to prove that $a_{ik}=a_{ki}=0$ for all $k \neq j$?
First assume that all eigenvalues of $ A=(a_{ij})$ are equal to $\lambda$ and that $diag A=\lambda I_n.$ Then $A=\lambda I_n.$ To see this we compute trace$ A^2$ in two ways and we get $\sum_{ij}|a_{ij}|^2=n\lambda^2$ which implies $|a_{ij}|^2=0$ for $i\neq j.$
Consider now the general case where the maximal eigenvalue $\lambda$ appears $n-k$ times exactly on the diagonal of $A$. Without loss of generality we assume that $A$ is written by blocks $(k,n-k)$ $$A=\left[\begin{array}{cc}A_1&B\\B^*&A_2\end{array}\right]$$ where $diag A_2=\lambda I_{n-k}.$ Let us prove that $B=0.$ For this we diagonalize $A$ with a unitary matrix $U$ : $$A=\left[\begin{array}{cc}U_1&V\\W^*&U_2\end{array}\right]\left[\begin{array}{cc}D&0\\0&\lambda I_{n-k}\end{array}\right]\left[\begin{array}{cc}U_1^*&W\\V^*&U^*_2\end{array}\right]$$ Therefore $$\lambda U_2U_2^*+W^*DW=\lambda I_{n-k}\stackrel{*}{=}\lambda (U_2U_2^*+W^*W)$$ (using in (*) the fact that $U$ is unitary). Hence $W^*(\lambda I_{n-k}-D)W=0$ which implies that $W=0$ since $\lambda I_{n-k}-D$ is positive definite. From the fact that $U$ is unitary we have $U_1W+VU_2^*=0$ . As a consequence $VU_2^*=0.$ Finally since $B=U_1DW+\lambda VU^*_2$ we get $B=0$ as desired.
Let us now prove that $A_2-\lambda I_{n-k}=0.$ This comes from the fact that $$A=\left[\begin{array}{cc}A_1&0\\0&A_2\end{array}\right]$$ and from the first remark of this post applied to $A_2.$ Finally $$A=\left[\begin{array}{cc}A_1&0\\0&\lambda I_{n-k}\end{array}\right].$$