Solution:

Henceforth, we assume that every equation has some $a_{i} \neq 0$, since otherwise it is of the form $0=a$, and then we can change it to either $0 \leq a$ or $0 \geq a$ to form an inequality which is always true. We use induction on $n$.

If $n=1$, we have the two equations $a x=b$, $c x=d$. We can divide out by $a$ and $c$ to obtain the two equations $x=b'$, $x=d'$ equivalent to our original equations. Without loss of generality we can assume $b' \leq d'$; then if we form the inequalities equivalent to $x \geq b'$, $x \leq d'$, at least one of these always holds.

Suppose $n>1$ and that the statement has been proven for $n$ equations in $n-1$ variables. Given $n+1$ equations in $n$ variables, look at the last equation. By assumption, some coefficient - say the coefficient of $x_{n}$ - is nonzero, so we can solve this equation for $x_{n}$ in terms of the other variables. Plugging this result into the first $n$ equations, we obtain $n$ linear equations in $x_{1}, \ldots, x_{n-1}$. By the induction hypothesis, we can replace each of these equalities with a $\leq$ or $\geq$ sign so that, for any choice of $x_{1}, \ldots, x_{n-1}$, some one of these $n$ inequalities is true.

We thus assign either the $\leq$ or the $\geq$ sign for the first $n$ original equations, and we claim we can choose the sign in our $n+1$th equation so that this result still holds. Suppose, otherwise, that, either way we choose the sign, some choice of the variables causes all of the inequalities to be violated. If the $n+1$th equation was $a_{1} x_{1}+\cdots+a_{n} x_{n}=a$, then we can choose $x_{1}', \ldots, x_{n}'$ to violate the first $n$ inequalities so that $a_{1} x_{1}'+\cdots+a_{n} x_{n}'=a'<a$, and we can also choose $x_{1}'', \ldots, x_{n}''$ to violate the first $n$ inequalities so that $a_{1} x_{1}''+\cdots+a_{n} x_{n}''=a''>a$. Now, consider what happens when we set
$$x_{i}=\frac{\left[a''-a\right] x_{i}'+\left[a-a'\right] x_{i}''}{a''-a'}.$$
The $n+1$th equation now holds (as an equality), since
$$\sum_{i=1}^{n} a_{i}\frac{\left[a''-a\right] x_{i}'+\left[a-a'\right] x_{i}''}{a''-a'}=\frac{\left[a''-a\right] \sum_{i=1}^{n} a_{i} x_{i}'+\left[a-a'\right] \sum_{i=1}^{n} a_{i} x_{i}''}{a''-a'}=\frac{\left[a''-a\right] a'+\left[a-a'\right] a''}{a''-a'}=a.$$
We claim that each of the first $n$ inequalities is still violated. Suppose, for example, that the inequality $c_{1} x_{1}+\cdots+c_{n} x_{n} \leq c$ is violated, so that $c_{1} x_{1}'+\cdots+c_{n} x_{n}'>c$ and $c_{1} x_{1}''+\cdots+c_{n} x_{n}''>c$. We can multiply the first of these inequalities by $a''-a$ and the second by $a-a'$ (preserving their signs, since these quantities are positive), add them, and divide by $a''-a'$ to obtain
$$\sum_{i=1}^{n} c_{i}\frac{\left[a''-a\right] x_{i}'+\left[a-a'\right] x_{i}''}{a''-a'}>c,$$
as claimed. The $\geq$ case is, of course, analogous. Thus, we have found values of $x_{i}$ for which the $n+1$th equality holds but none of the first $n$ inequalities hold. But this contradicts the way the first $n$ signs were chosen. So our assumption was wrong, and it was indeed possible to choose the sign of the $n+1$th inequality so that there would always be at least one true inequality, completing the induction step and the proof.