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$0 \ >\ \dfrac{1}{1-\frac{10}{x}}\ >\ 1 - \dfrac{5}{y}$

So...

$\dfrac{1}{1-\frac{10}{x}}\ >\ 1 - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}\ >\ 1 - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}-1\ >\ - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}-\dfrac{x-10}{x-10}\ >\ - \dfrac{5}{y}\\~\\~\\ \dfrac{10}{x-10}\ >\ -\dfrac5y$

And we know  0 < x < 10  because that is the only way  $\dfrac{1}{1-\frac{10}{x}}$  can be less than 0.

So  x - 10  is negative, and so when we multiply both sides by  (x - 10) ,  flip the sign.

$10\ <\ -\dfrac5y(x-10)$

And we know  0 < y < 5  because that is the only way  $1-\dfrac5y$  can be less than 0.

So  y  is positive, and so when we multiply both sides by  y ,  don't flip the sign.

$10y\ <\ -5(x-10)\\~\\ y\ <\ -\frac12(x-10)\\~\\ y\ <\ -\frac12x+5$

And so we have these three inequalities:

$1.\qquad0\ {<}\ x\ {<}\ 10\\~\\ 2.\qquad0\ {<}\ y\ {<}\ 5\\~\\ 3.\qquad y\ {<}\ -\frac12x+5$

We can see on a graph that the intersection of these is a triangle:

https://www.desmos.com/calculator/2xloyva03v

All the points that lie within the shaded triangle are solutions to the inequality.

And we can see there are 16 pairs of integers (x, y) that satisfy the inequality.