Prove that $\sum\limits_{i=1}^{n} a_i\geq n^2$.

Your conjecture is true.

If you want only a hint, stop reading after recalling that $\gcd(a_i,a_j) \mid i-j$ is equivalent to the solvability of the system of congruences $$ \left\{ \begin{align*} x \equiv i \pmod {a_i} \\ x \equiv j \pmod {a_j}\end{align*}\right.$$

By the Chinese remainder theorem, we can formulate the problem equivalently as follows: consider $n$ arithmetic progressions in $\mathbb Z$, $(S_i)_{1 \leq i \leq n}$ with $i \in S_i$ for all $i$ and with differences $a_i$. That is, $S_i = i + a_i \mathbb Z$. If the $S_i$ are pairwise disjoint, prove that $\sum a_i \geq n^2$.

The conclusion remains true even without the hypothesis that $i \in S_i$:

If the $S_i$ are disjoint, they are also disjoint mod $a_1 \cdots a_n$, and hence $$\begin{align*} \sum_{i=1}^n\# (S_i \bmod a_1 \cdots a_n) & \leq a_1 \cdots a_n \\ \sum_{i=1}^n \prod_{j \neq i} a_j &\leq \prod_{i=1}^n a_i \end{align*}$$ divide by the RHS to obtain $$\sum_{i=1}^n\frac1{a_i} \leq 1$$ And by AM-HM $$\sum_{i=1}^n a_i \geq n^2$$

Also from AM-HM we have that equality occurs iff $a_1 = \cdots = a_n = n$.