Claim 1:$\sum_{i=1}^{n} a_i \leq n^2+2025\quad$ ProofIt is a result of Cauchy-Schwartz inequality. $(\sum_{i=1}^{n} a_i )^2 \leq n.\sum_{i=1}^{n} a_i^2 =n^4+2025n^2\leq (n^2+2025)^2$ Claim 2:$a_n \geq 2n-2026$ Proof$a_{n}=\sum_{i=1}^{n} a_i-\sum_{i=1}^{n-1} a_i \geq n^2-(n-1)^2-2025=2n-2026$ There doesn't exist such sequence. Contradiction$n^3+2025n \geq \sum_{i=1}^{n} a_i^2 \geq \sum_{i=1013}^{n} a_i^2 \geq \sum_{i=1013}^{n} (2i-2026)^2 =4.\sum_{i=0}^{n-2013} (i)^2= 4.\frac{(n-2013).(n-2012).(2n-2015)}{6}$ Leading coefficient $[n^3]$ is $1$ at the left handside whereas it is $\frac{4}{3}$ at the right handside which is a contradiction for large values of $n$.