Find the smallest real constant $C$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that  $$ \left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C . $$  Answer. The smallest $C$ is $\frac{1}{2}$.
We first show that \( C \leqslant \frac{1}{2} \). For any positive real numbers \( a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5} \), consider the five fractions 
\[ \frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, \frac{a_{1}}{a_{5}}, \frac{a_{2}}{a_{3}}, \frac{a_{4}}{a_{5}}. \]
Each of them lies in the interval \((0,1]\). Therefore, by the Pigeonhole Principle, at least three of them must lie in \(\left(0, \frac{1}{2}\right]\) or lie in \(\left(\frac{1}{2}, 1\right]\) simultaneously. In particular, there must be two consecutive terms in (2) which belong to an interval of length \(\frac{1}{2}\) (here, we regard \(\frac{a_{1}}{a_{2}}\) and \(\frac{a_{4}}{a_{5}}\) as consecutive). In other words, the difference of these two fractions is less than \(\frac{1}{2}\). As the indices involved in these two fractions are distinct, we can choose them to be \(i, j, k, l\) and conclude that \( C \leqslant \frac{1}{2} \).

Next, we show that \( C = \frac{1}{2} \) is best possible. Consider the numbers \(1, 2, 2, 2, n\) where \(n\) is a large real number. The fractions formed by two of these numbers in ascending order are \(\frac{1}{n}, \frac{2}{n}, \frac{1}{2}, \frac{2}{2}, \frac{2}{1}, \frac{n}{2}, \frac{n}{1}\). Since the indices \(i, j, k, l\) are distinct, \(\frac{1}{n}\) and \(\frac{2}{n}\) cannot be chosen simultaneously. Therefore the minimum value of the left-hand side of (1) is \(\frac{1}{2} - \frac{2}{n}\). When \(n\) tends to infinity, this value approaches \(\frac{1}{2}\), and so \(C\) cannot be less than \(\frac{1}{2}\).

These conclude that \(C = \frac{1}{2}\) is the smallest possible choice. Comment. The conclusion still holds if \(a_{1}, a_{2}, \ldots, a_{5}\) are pairwise distinct, since in the construction, we may replace the 2's by real numbers sufficiently close to 2. There are two possible simplifications for this problem: (i) the answer \(C = \frac{1}{2}\) is given to the contestants; or (ii) simply ask the contestants to prove the inequality (1) for \(C = \frac{1}{2}\).