3. $n \in(-1,1)$.
(1) When $n \leqslant-1$, by
$$
\begin{array}{l}
x0 \Rightarrow m>y,
\end{array}
$$

it leads to a contradiction.
(2) When $n \geqslant 1$, by
$$
n+1 \geqslant y=m+x>2 x \geqslant 2 n \Rightarrow n<1,
$$
it leads to a contradiction.
(3) When $-1<n<1$,
if $-1<n<\frac{1}{2}$, take $y=n+1, m=\frac{2(n+1)}{3}$, $x=\frac{n+1}{3}$;
if $\frac{1}{2} \leqslant n<1$, take $y=n+1, m=1, x=n$.
In summary, $n \in(-1,1)$.
3. Given real numbers $x, y$ satisfy $n \leqslant x<y \leqslant n+1$. If there exist $x, y$ and $m \in(x, y)$, such that $|y|=|m|+|x|$, $m, n$ are real numbers, then the range of values for $n$ is $\qquad$ .