Denote by $ N(e)$ the number assigned to any edge $ e\in E(G).$ The solution is based on the observation, the vertices with large degrees are not too many. Namely $$\displaystyle \left|\{v\in V(G): \deg(v)\ge d\}\right|\le \frac{ 2|E(G)| }{d} \quad (1)$$Take an edge with number at least $ d$ written on it. Then both of its ends must have degrees at least $ d$. By (1) we get $$ \displaystyle |\{e\in E(G): N(e)\ge d\}|\le \frac{1}{2}|\{v\in V(G): \deg(v)\ge d\}|^2 \leq \frac{2|E(G)|^2}{d^2}\quad (2)$$Denote $$ \displaystyle D(d_1,d_2):= \sum_{e\in E(G), d_1\le N(e)< d_2} N(e)$$We want to estimate $ D(1,n)$. Clearly $$ \displaystyle D(1,\sqrt{n})\le |E(G)|\sqrt{n}=n\sqrt{n}\quad (3)$$We estimate $ D(\sqrt{n},n)$ as follows $$ \displaystyle D(\sqrt{n},n)=\sum_{j=1}^k D(2^{j-1}\sqrt{n},2^{j}\sqrt{n})$$where $ k:=\left\lceil\log\sqrt{n}\right\rceil$. Obviously $$ \displaystyle D(2^{j-1}\sqrt{n},2^{j}\sqrt{n})\le |\{e\in E(G): N(e)\ge 2^{j-1}\sqrt{n}\}| 2^j \sqrt{n}$$Hence, by (2) we get $$ \displaystyle D(2^{j-1}\sqrt{n},2^{j}\sqrt{n}) \le \frac{2n^2}{2^{2(j-1)}n}\cdot 2^j\sqrt{n}= 2^{3-j}n\sqrt{n}$$$$ \displaystyle D(\sqrt{n},n) \le n\sqrt{n}\sum_{j=1}^k 2^{3-j}<8n\sqrt{n}$$Taking into account (3) finally we obtain $$ \displaystyle D(1,n)\le 9n\sqrt{n}$$This estimate is sharp (up to multiplication by a constant). Indeed when $ G$ is a complete or almost complete graph, $ D(G)\approx \sqrt{2}n\sqrt{n}$. Remark. I put it also in my blog, where another possible approaches have been considered.