Cute. For $n = 28$ it works: take $\alpha(i) = \beta(i) = \gamma(i) = i$ and $\delta(i) = 29 - i$, then we get \[ \sum_{i = 1}^{28}\alpha(i)\beta(i) = \sum_{i = 1}^{28}i^2 = 7714 \] and \[ \frac{19}{10} \cdot \sum_{i = 1}^{28}\gamma(i)\delta(i) = \frac{19}{10} \cdot \sum_{i = 1}^{28}i(29 - i) = \frac{19}{10} \cdot 4060 = 7714. \] However, it cannot work for $n \leq 27$. Indeed, by the rearrangement inequality we have \[ \frac{19}{10} \cdot \sum_{i = 1}^n\gamma(i)\delta(i) \geq \frac{19}{10} \cdot \sum_{i = 1}^ni(n + 1 - i) = \frac{19}{10} \cdot \frac{n(n + 1)(n + 2)}{6} \] and \[ \sum_{i = 1}^n\alpha(i)\beta(i) \leq \sum_{i = 1}^ni^2 = \frac{n(n + 1)(2n + 1)}{6}. \] Now, since $n \leq 27$, \[ \frac{19}{10} \cdot (n + 2) > 2n + 1 \] (since this is equivalent to $n < 28$), so the right-hand side will always be "too big". So, the solution is $n = 28$.