Thank you Peter! The matrix is symmetric.

Thank you Peter for the idea! Do we imply that all the diagonal entries are equal to 1 or only the maximal of them?

But in your example the sum of off-diagonal elements of the matrix is 4 and the sum of diagonal elements is 9 so it is not a counterexample. In general if all the diagonal elements are equal then the above claim is true.

For the case of non-symmetric matrix the example is (0 0 0; 1 1 1; 0 0 0) (later: this is not a good example since this matrix is not a positive semidefinite) but what regarding the symmetric case?

Thank you Peter, at least you outlined several good ideas.

Thank you very much Peter! This question is interesting to me because I currently research one problem and this bound may be of use for me. May be someone will read our discussion and it will encourage him to use this question in the process of preparation students. Anyway we do a good job for the coming generations :)

Thank you Peter! I'll try to understand what you written in the nearest days.

 All right Peter, thank you for the help!

The use of complement is smart.

I hope that everything goes normally in your university!

The counterexample is as follows:

a(i,j) = 1/(ij)

Clearly, this matrix is of the form v v^T and is nonnegative definite.

For the 1st row, the claim now reads as 

sum_{i=2}^n 1/j  

Thank you!

I am sorry for the misprint: the example must be read as

sum_{i=2}^n 1/i < sum_{i=1}^n 1/i^2

In general, a good (unimprovable) bound follows if a(i,i)=1 for all i; in this case

sum_{i \= j} |a(i,j)|

Thank you!

A Hermitian diagonally dominant matrix  with real non-negative diagonal entries is positive semidefinite. 

Please see a proof in the attachment:

https://en.wikipedia.org/wiki/Diagonally_dominant_matrix

Thank you Wiwat Wanicharpichat!