Denote $b_i=a_1+a_2+\dots+a_i\,,\,i= 1,2,\dots,n$. Arguing by contradiction, assume there are no more than $\lfloor \sqrt{n}\rfloor$ different $b_i$'s $\pmod{n}$. Jumping from $b_1$ to $b_2$ , ... to $b_n$, all possible differences $b_{i+1}-b_i$ can't be more than $\lfloor \sqrt{n}\rfloor(\lfloor\sqrt{n}\rfloor -1)+1$, since only one time we will jump from one residue to the same one. Thus, the possible values of $a_i=b_i-b_{i-1}$ will be less than $n$, a contradiction.

huricane wrote: Let $n\ge 5$ be a positive integer and let $\{a_1,a_2,...,a_n\}=\{1,2,...,n\}$.Prove that at least $\lfloor \sqrt{n}\rfloor +1$ numbers from $a_1,a_1+a_2,...,a_1+a_2+...+a_n$ leave different residues when divided by $n$. Kvant 1991 : http://kvant.ras.ru/1991/07/resheniya_zadach_m1266_-_m1270.htm