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This question could be asked on MSE.

Using the rewording of the problem by Gerry Myerson, the analysis below shows it is not possible for some $n$. For $n=2$ it is possible, but for $n\geq 3$ it is not. Take the case $n=3$, and let the entries of the first vector be denoted $a,b,c$ with those in the second vector $d,e,f$, so that outer-product matrix is given by $$\begin{pmatrix} ad & ae & af \\ bd & be & bf \\ cd & ce & cf \end{pmatrix}.$$

The conditions mandate that $ad\geq1$, $be\geq1$, so giving $adbe\geq 1$, and $ae<1$, $bd<1$ give $ae$ and $bd$ are negative. Similar considerations imply $af,bf,cd,ce$ are negative. However, it must then be the case that exactly $a$ or $e$ is negative, and the same is true for $b$ and $d$, $c$ and $d$, and so on. If $a$ and $d$ are negative then $e$ is positive and so $ce$ is positive, contradicting the assumptions. But then $d$ is positive, so that $c$ and $b$ are negative. Also, $e$ is positive (since $c$ is negative). Alas this is a problem, as $be$ is now negative. Changing the inequalities from '$\geq$' to '$>$' does not remedy this problem.

(Any matrix of size larger than $n=3$ will have a $n=3$ system as a submatrix so it suffices to consider $n=3$ to answer the question of existence for larger $n$.)

This question could be asked on MSE.

Using the rewording of the problem by Gerry Myerson, the analysis below shows it is not possible for some $n$. For $n=2$ it is possible, but for $n\geq 3$ it is not. Take the case $n=3$, and let the entries of the first vector be denoted $a,b,c$, with those in the second vector $d,e,f$, so that outer-product matrix is given by $$\begin{pmatrix} ad & ae & af \\ bd & be & bf \\ cd & ce & cf \end{pmatrix}.$$

The conditions mandate that $ad\geq1$, $be\geq1$, so giving $adbe\geq 1$, and $ae<1$, $bd<1$ give $ae$ and $bd$ are negative. Similar considerations imply $af,bf,cd,ce$ are negative. However, it must then be the case that exactly $a$ or $e$ is negative, and the same is true for $b$ and $d$, $c$ and $d$, and so on. If $a$ and $d$ are negative then $e$ is positive and so $ce$ is positive, contradicting the assumptions. But then $d$ is positive, so that $c$ and $b$ are negative. Also, $e$ is positive (since $c$ is negative). Alas this is a problem, as $be$ is now negative. Changing the inequalities from '$\geq$' to '$>$' does not remedy this problem.

(Any matrix of size larger than $n=3$ will have a $n=3$ system as a submatrix so it suffices to consider $n=3$ to answer the question of existence for larger $n$.)

This question could be asked on MSE.

Using the rewording of the problem by Gerry Myerson, it is not possible for some $n$. For $n=2$ it is possible, but for $n\geq 3$ it is not. Take the case $n=3$, $$\begin{pmatrix} ad & ae & af \\ bd & be & bf \\ cd & ce & cf \end{pmatrix}.$$

The conditions mandate that $ad\geq1$, $be\geq1$, so giving $adbe\geq 1$, and $ae<1$, $bd<1$ give $ae$ and $bd$ are negative. Similar considerations imply $af,bf,cd,ce$ are negative. However, it must then be the case that exactly $a$ or $e$ is negative, and the same is true for $b$ and $d$, $c$ and $d$, and so on. If $a$ and $d$ are negative then $e$ is positive and so $ce$ is positive, contradicting the assumptions. But then $d$ is positive, so that $c$ and $b$ are negative. Also, $e$ is positive (since $c$ is negative). Alas this is a problem, as $be$ is now negative. Changing the inequalities from '$\geq$' to '$>$' does not remedy this problem.

(Any matrix of size larger than $n=3$ will have a $n=3$ system as a submatrix so it suffices to consider $n=3$ to answer the question of existence for larger $n$.)

This question could be asked on MSE.

Using the rewording of the problem by Gerry Myerson, the analysis below shows it is not possible for some $n$. For $n=2$ it is possible, but for $n\geq 3$ it is not. Take the case $n=3$, and let the entries of the first vector be denoted $a,b,c$ with those in the second vector $d,e,f$, so that outer-product matrix is given by $$\begin{pmatrix} ad & ae & af \\ bd & be & bf \\ cd & ce & cf \end{pmatrix}.$$

The conditions mandate that $ad\geq1$, $be\geq1$, so giving $adbe\geq 1$, and $ae<1$, $bd<1$ give $ae$ and $bd$ are negative. Similar considerations imply $af,bf,cd,ce$ are negative. However, it must then be the case that exactly $a$ or $e$ is negative, and the same is true for $b$ and $d$, $c$ and $d$, and so on. If $a$ and $d$ are negative then $e$ is positive and so $ce$ is positive, contradicting the assumptions. But then $d$ is positive, so that $c$ and $b$ are negative. Also, $e$ is positive (since $c$ is negative). Alas this is a problem, as $be$ is now negative. Changing the inequalities from '$\geq$' to '$>$' does not remedy this problem.

(Any matrix of size larger than $n=3$ will have a $n=3$ system as a submatrix so it suffices to consider $n=3$ to answer the question of existence for larger $n$.)

This question is better asked on MSE most likely.

Using the rewording of the problem by Gerry Myerson, it is not possible for some $n$. For $n=2$ it is possible, but for $n\geq 3$ it is not. Take the case $n=3$, $$\begin{pmatrix} ad & ae & af \\ bd & be & bf \\ cd & ce & cf \end{pmatrix}.$$

The conditions mandate that $ad\geq1$, $be\geq1$, so giving $adbe\geq 1$, and $ae<1$, $bd<1$ give $ae$ and $bd$ are negative. Similar considerations imply $af,bf,cd,ce$ are negative. However, it must then be the case that exactly $a$ or $e$ is negative, and the same is true for $b$ and $d$, $c$ and $d$, and so on. If $a$ is negative and $d$ is then $e$ is positive and so $ce$ is positive, contradicting the assumptions. But then $d$ is positive, so that $c$ and $b$ are negative. Also, $e$ is positive (since $c$ is negative). Alas this is a problem, as $be$ is now negative. Changing the inequalities from '$\geq$' to '$>$' does not remedy this problem.

(Any matrix of size larger than $n=3$ will have a $n=3$ system as a submatrix so it suffices to consider $n=3$ to answer the question of existence for larger $n$.)

This question could be asked on MSE.

Using the rewording of the problem by Gerry Myerson, it is not possible for some $n$. For $n=2$ it is possible, but for $n\geq 3$ it is not. Take the case $n=3$, $$\begin{pmatrix} ad & ae & af \\ bd & be & bf \\ cd & ce & cf \end{pmatrix}.$$

The conditions mandate that $ad\geq1$, $be\geq1$, so giving $adbe\geq 1$, and $ae<1$, $bd<1$ give $ae$ and $bd$ are negative. Similar considerations imply $af,bf,cd,ce$ are negative. However, it must then be the case that exactly $a$ or $e$ is negative, and the same is true for $b$ and $d$, $c$ and $d$, and so on. If $a$ and $d$ are negative then $e$ is positive and so $ce$ is positive, contradicting the assumptions. But then $d$ is positive, so that $c$ and $b$ are negative. Also, $e$ is positive (since $c$ is negative). Alas this is a problem, as $be$ is now negative. Changing the inequalities from '$\geq$' to '$>$' does not remedy this problem.

(Any matrix of size larger than $n=3$ will have a $n=3$ system as a submatrix so it suffices to consider $n=3$ to answer the question of existence for larger $n$.)