18. 1.5 * Find all positive integer solutions \(x, y, z, t\) to the equation \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1\) with \(x \leqslant y \leqslant z \leqslant t\).
If the equation has a solution, then it must be that $x \leqslant 4$. Because when $x \geqslant 5$, due to $x \leqslant y \leqslant z \leqslant t$, we have $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} \leqslant \frac{4}{5} \frac{1}{z}$, so at this time, the possible values of $z$ are $5, 6, 7, 8$. When $z = 5, 6, 7, 8$, we respectively get $t$ as $20, 12, \frac{28}{3}, 8$. Thus, we obtain three solutions of the equation: $x=2, y=4, z=5, t=20; x=2, y=4, z=6, t=12; x=2, y=4, z=8, t=8$.

If $y=5$, then $\frac{3}{10} > \frac{1}{z} + \frac{1}{t} \leqslant \frac{2}{z}$, which gives $z \leqslant 6$, and at the same time $z \geqslant y=5$. So at this time, $z$ can only be 5 or 6. When $z$ is 5 or 6, we respectively get $t$ as $10, \frac{15}{2}$. Thus, we obtain one solution of the equation: $x=2, y=5, z=5, t=10$.

If $y=6$, then $\frac{1}{3} = \frac{1}{z} + \frac{1}{t} \leqslant \frac{2}{z}$, which gives $z \leqslant 6$, and also $z \geqslant y=6$. So $z=6$. At this time, $t=6$. Thus, $x=2, y=6, z=6, t=6$ is a solution of the equation.

Similarly to the case when $x=2$, consider the cases when $x=3, 4$. We can obtain 14 solutions of the equation where $x \leqslant y \leqslant z \leqslant t$, namely $2,3,7,42; 2,3,8,24; 2,3,9,18; 2,3,10,15; 2,3,12,12; 2,4,5,20; 2,4,6,12; 2,4,8,8; 2,5,5,10; 2,6,6,6; 3,3,4,12; 3,3,6,6; 3,4,4,6; 4,4,4,4$.