Prove that $\lim_{n\rightarrow\infty}\frac{x_1^2+x_2^2+\cdots+x_n^2}{n^2}=0$

Your first requirement implies that there is some $R\in\left(0,\infty\right)$ such that $$ \sum_{i=1}^{n}\frac{x_{i}}{n}\leq R $$ holds for all $n\in\mathbb{N}$.

Now, let $\varepsilon>0$ be arbitrary. Your second requirement shows that there is some $n_{1}\in\mathbb{N}$ such that $\frac{x_{n}}{n}<\frac{\varepsilon}{2R}$ holds for all $n\geq n_{1}$.

Let $S:=\max_{i=1,\dots,n_{1}}x_{i}$ and choose $n_{2}\in\mathbb{N}$ with $n_{2}\geq n_{1}$ and $\frac{SR}{n_{2}}\leq\varepsilon/2$.

We now derive for $n>n_{2}\geq n_{1}$ that \begin{eqnarray*} \sum_{i=1}^{n}\frac{x_{i}^{2}}{n^{2}} & = & \left[\max_{i=1,\dots,n_{1}}x_{i}\right]\cdot\frac{1}{n}\cdot\sum_{i=1}^{n_{1}}\frac{x_{i}}{n}+\left[\max_{j=n_{1}+1,\dots,n}\frac{x_{j}}{n}\right]\cdot\sum_{i=n_{1}+1}^{n}\frac{x_{i}}{n}. \end{eqnarray*}

I believe you can take it from there. Otherwise:

\begin{eqnarray*}\dots \\ & \leq & \frac{SR}{n}+\left[\max_{j=n_{1}+1,\dots,n}\frac{x_{j}}{j}\right]\cdot\sum_{i=1}^{n}\frac{x_{i}}{n}\\ & \leq & \frac{SR}{n}+\frac{\varepsilon}{2R}\cdot R\leq\varepsilon.\end{eqnarray*}

Remark: What we are doing in the above is essentially an interpolation argument between an $\ell^1$ estimate and an $\ell^\infty$ estimate to get an $\ell^2$ estimate . The only problem is that the $\ell^\infty$ estimate only holds in the limit, so that one has to treat the first finitely many terms in a special way.