Let's divide 17 into two parts such that $3/4$ of one part is $a$ greater than $5/6$ of the other part, where $a>0$. What are the limits within which the value of $a$ can vary to ensure that the problem is always solvable?
One part is denoted by $x$, the other by $17-x$, and the conditional equation is

$$
\frac{3}{4} x=\frac{5}{6}(17-x)+a
$$

Rearranging the equation, we get

$$
x=\frac{170+12 a}{19}
$$

From the condition, it follows that $x$ is greater than 0 and less than 17, that is,

$$
0<\frac{170+12 a}{19}<17
$$

The first condition is satisfied for all positive $a$, and the second condition holds if

$$
170+12 a<17 \cdot 19
$$

or

$$
4 a<51 ; \quad a<12 \frac{3}{4}
$$

Solved by: Dömölki B., Főző Éva, Kántor S., Kovács L., Szabó Magda, Zobor E.