3 Given a set of positive real numbers $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ satisfying the inequalities
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}<\frac{1}{2}\left(x_{1}+x_{2}+\cdots+x_{k}\right), \\
x_{1}+x_{2}+\cdots+x_{k}<\frac{1}{2}\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}\right),
\end{array}
$$

try to find the smallest value of $k$ that satisfies the conditions.
The smallest value of $k$ that satisfies the conditions is 516.
(1) Taking $x_{1}=5.169, x_{2}=x_{3}=\cdots=x_{516}=0.125$ satisfies the two inequalities in the problem.
(2) We now prove: $k \geqslant 516$.
Let $x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2} > 0$) we get
$$
x_{1}>4 \text{. }
$$

Thus, we can assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{s}>4(s \geqslant 1)$, and without loss of generality,
$$
\frac{1}{4} \geqslant x_{s+1} \geqslant x_{s+2} \geqslant \cdots \geqslant x_{k}
$$
(In fact, if $x_{s+1} \geqslant \frac{1}{2}$, we can directly remove it, which reduces the value of $k$ and the difference on both sides of (1) and (2) decreases, so (1) and (2) still hold; if $\frac{1}{4}s$, otherwise (2) does not hold. Let $t=k-s(t \geqslant 1, t \in \mathbf{N}$.

Let $f_{1}(x)=x^{2}-\frac{1}{2} x\left(00, f_{2}^{\prime \prime}(x)=6 x-8t\left(4 x_{0}^{2}-x_{0}^{3}\right) .
$$

Since $00, 4 x_{0}^{2}-x_{0}^{3}>0$, and from (4) and (5) we get
$$
\frac{\sum_{i=1}^{1}\left(2 x_{i}^{2}-x_{i}\right)}{x_{0}-2 x_{0}^{2}}0 \text{, }
$$

therefore there exists $r \in\{1,2, \cdots, s\}$ such that
$$
\frac{x_{r}^{3}-4 x_{r}^{2}}{4 x_{0}^{2}-x_{0}^{3}}-\frac{2 x_{r}^{2}-x_{r}}{x_{0}-2 x_{0}^{2}}>0 .
$$

Simplifying, we get $\left(1-2 x_{0}\right)\left(2 x_{r}-1\right)>7$, so
$$
x_{r}>\frac{4-x_{0}}{1-2 x_{0}} \text{. }
$$

Noting that for $i=1,2, \cdots, s, 2 x_{i}^{2}-x_{i}>0$, and $\frac{4-x_{0}}{1-2 x_{0}}>4$, hence
$$
\begin{aligned}
t & >\frac{\sum_{i=1}^{3}\left(2 x_{i}^{2}-x_{i}\right)}{x_{0}-2 x_{0}^{2}}>\frac{2 x_{r}^{2}-x_{r}}{x_{0}-2 x_{0}^{2}}>\frac{2\left(\frac{4-x_{0}}{1-2 x_{0}}\right)^{2}-\frac{4-x_{0}}{1-2 x_{0}}}{x_{0}-2 x_{0}^{2}} \\
& =\frac{7\left(4-x_{0}\right)}{\left(1-2 x_{0}\right)^{3} x_{0}} .
\end{aligned}
$$

Let $y=1-2 x 0+$ and let $g(y)=\frac{7\left(4-\frac{1-y}{2}\right)}{y^{3} \cdot \frac{1-y}{2}}$. Using the method of derivatives, it is not difficult to find that the minimum value of $g(y)$ is between 514 and 515, thus $t \geqslant 515$, so $k \geqslant t+$ $1 \geqslant 516$.
In summary, the smallest value of $k$ that satisfies the conditions is 516.