Key idea.

Use the positivity condition to bound the smaller number.

Write the two numbers as \(y\) and \(2y+k\), where \(y\) and \(k\) are positive integers. Their sum gives

Since \(k>0\), we need \(3y<28\). Thus \(y\) can be any integer from \(1\) to \(9\).

These choices give the distinct positive values

For every choice, both original numbers are positive. Hence there are exactly \(9\) possible values.