Positive-definiteness of matrix with $|a_{ij}| \leq \frac{1}{ij}$ Let $A$ be a $N\times N$-matrix with elements 
$$
a_{ii}=1 \quad\text{and}\quad
a_{ij} = \frac{1}{ij}  \quad\text{for}~ i\neq j.
$$
Then $A$ is positive-definite, as can be easily seen from
$$
x^T A x = \sum_i x_i^2 + \sum_{i \neq j} \frac{x_i x_j}{ij} 
\geq
\sum_i \frac{x_i^2}{i^2} + \sum_{i \neq j} \frac{x_i x_j}{ij} 
=
\left(\sum_i \frac{x_i}{i}\right)^2 \geq 0.
$$
Assume now that $A$ is a real symmetric $N\times N$-matrix with elements 
$$
\tag{1}
a_{ii}=1 \quad\text{and}\quad
|a_{ij}| \leq \frac{1}{ij}  \quad\text{for}~ i\neq j.
$$

Is it possible to show that $A$ is also positive-definite (or positive-semidefinite)? 


Here I posted a refinement of this question by assuming $0 \leq a_{ij} \leq \frac{1}{ij}$ for $i \neq j$ in $(1)$.
 A: It's false.
Choose $N=20$ and, for every $i\not= j$, $A_{i,j}=\dfrac{-1}{ij}$. $A$ admits a negative eigenvalue $\approx -0.04$.
It suffices to consider your first matrix $A$ -denoted by $A0$ (with $A0_{i,j}=1/(ij)$)-.
After, increase $N$ until $\rho(A0)> 2$.
A: Precise computations dealing with the solution by loup blanc:
Let 
$$A_N=\begin{pmatrix}
1&-1/2&-1/3&\cdots&-1/N\\
-1/2&1&-1/6&\cdots&-1/(2N)\\
-1/3&-1/6&1&\cdots&-1/(3N)\\
\cdots&\cdots&\cdots&\cdots&\cdots\\
-1/N&-1/(2N)&-1/(3N)&\cdots&1\\
\end{pmatrix}$$
(diagonal elements equal to $1$, off-diagonal elements $(A_N)_{ij}=-\dfrac{1}{ij}$.)
We are going to obtain a formula for $\det(A_N)$ (formula (2)) showing a certain value $N_0$ exists such that 
$$\text{for all} \ N \geq N_0, \ \ \det(A_{N})<0$$ 
Let us use the following formula for the determinant of an invertible matrix $M$ and (column) vectors $U,V \in \mathbb{R^n}$
$$\det(M+UV^T)=\det(M)(1+V^TM^{-1}U)\tag{1}$$
(matrix-determinant lemma)
If we take :
$$\begin{cases}M&=&diag(1+1/1^2,1+1/2^2,1+1/3^2, \cdots 1+1/N^2)\\
U&=&(1,1/2,1/3,\cdots 1/N)^T\\
V&=&-U\end{cases},$$
we have :
$$A_N=M+UV^T.$$ 
Therefore, using (1) :
$$\det(A_N)=\prod_{k=1}^N \left(1+\dfrac{1}{k^2}\right) \times \left(1-\sum_{k=1}^N \dfrac{1}{k} \times \dfrac{1}{1+\tfrac{1}{k^2}} \times \dfrac{1}{k}\right)$$
$$\det(A_N)=\prod_{k=1}^N \left(1+\dfrac{1}{k^2}\right) \times \left(1-\sum_{k=1}^N \dfrac{1}{k^2+1}\right)\tag{2}$$
But the following series is convergent with sum :
$$\sum_{k=1}^{+\infty} \dfrac{1}{k^2+1}=\dfrac12\left(\pi \coth \pi -1 \right)\approx 1.07667...> 1$$
(see a proof here or there).
Therefore, the content of the second parenthesis in (2) becomes negative beyond a certain $N_0$ as expected ; thus $\det(A_N)<0 $ for $N \geq N_0$ : $A_N$ cannot be semi-definite positive for  $N \geq N_0$.
This value $N_0$ happens to be $13$ as "forecasted" by numerical attempts.
Remarks : 
1) The first part of the determinant above tends, when $N \to \infty$ to the following convergent infinite product (see this) :
$$\prod_{k=1}^{\infty} \left(1+\dfrac{1}{k^2}\right)=\dfrac{\text{sinh} \ \pi}{\pi}\approx 3.676078.$$
As a consequence, for very large values of $N$, 
$$\det(A_N) \approx \dfrac{1}{2}\left(\dfrac{1}{\pi}(3 \sinh \pi)- \cosh \pi\right) = -0.28186..$$
2) It is instructive to plot the eigenvalues of $A_N$ (which are all real because $A_N$ is symmetrical). See figure (horizontal axis : values of $N$ from $N=1$ to $N=200$). Different aspects can be underlined, in particular the fact that there is a unique negative eigenvalue ($\approx -0.10720$  for $N$ sufficiently large), a huge cluster just above $1$, and separate values at $\approx 1.0858226$, $\approx 1.176735$, and $\approx 1.593347$.
These curves are either increasing or decreasing as a consequence of Cauchy interlacing theorem for symmetric matrices.

A possible connection :(https://www.sciencedirect.com/science/article/pii/S002437951200198X).
