About fractions whose sum is a natural number

Some days ago I found an old problem of an olympiad that I always found interesting. It asks to replace each $\overline{){\displaystyle}}$ with the numbers $1,2\dots 30$ without repeating any number, such that their sum is an integer number.

$\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}$

I got a solution by trial-and-error and it's:

$\frac{14}{1}+\frac{23}{2}+\frac{11}{22}+\frac{19}{3}+\frac{10}{15}+\frac{29}{4}+\frac{9}{12}+\frac{17}{6}+\frac{5}{30}+\frac{25}{8}+\frac{21}{24}+\frac{16}{7}+\frac{20}{28}+\frac{27}{18}+\frac{13}{26}=53.$

I was wondering if there is another method different than trial-and-error. Thanks in advance for your answers.