Equation (iii) is equivalent to $(iv) \;\; \sum_{i=1}^n \frac{1}{a_i} = \frac{n}{2} - 1 + \frac{1}{15^{25} + 1}$. From the conditions (i) and (ii) we obtain $(v) \;\; a_i < 11 \;\;\; \Rightarrow \;\;\; a_i \in \{4,8\}$. Hence by equation (iv) $\frac{n}{4} \geq \sum_{i=1}^n \frac{1}{a_i} = \frac{n}{2} - 1 + \frac{1}{15^{25} + 1} > \frac{n}{2} - 1$, yielding ${\textstyle \frac{n}{4} < 1}$. Hence $n < 4$. Clearly $n>1$ by equation (iv). Consequently $n \in \{2,3\}$. If $n=2$, then according to equation (iv) $(vi) \;\; \sum_{i=1}^2 \frac{15^{25} + 1}{a_i} = 1$, which is impossible since the LHS of equation (vi) is $\geq 2$ by condition (ii). Therefore $n=3$, which inserted in equation (iv) result in $(vii) \;\; \sum_{i=1}^3 \frac{1}{a_i} = \frac{1}{2} + \frac{1}{15^{25} + 1}$. If $a_2 \neq 4$, then by implication (v) $\frac{1}{2} < \sum_{i=1}^3 \frac{1}{a_i} \leq \frac{3}{8}$, a contradiction implying $a_2=4$. Thus $a_1=4$ by conditions (i) and (ii). Setting $a_1=a_2=4$ in equation (vii), we obtain $a_3=15^{25} + 1$. Conclusion: The only solution of equation (iii) satisfying conditions (i) and (ii) is $n=3$ with $(a_1,a_2,a_3) = (4,4,15^{25}+1)$.