One. (20 points) Given the equations about $x$: $2 x^{2}-4 n x$ $-2 n=1$ and $x^{2}-(3 n-1) x+2 n^{2}-3 n=2$. Does there exist a value of $n$ such that the sum of the squares of the two real roots of the first equation equals an integer root of the second equation? If it exists, find such $n$ values; if not, explain the reason.
Given $\Delta_{1}=16 n^{2}-8(-2 n-1)$
$$
=16 n^{2}+16 n+8=(4 n+2)^{2}+4>0
$$

we know that for any real number $n$, the first equation has real roots.
Let the two roots of the first equation be $x_{1}$ and $x_{2}$, then we have
$$
x_{1}+x_{2}=2 n, x_{1} x_{2}=-\frac{2 n+1}{2} \text {. }
$$

Thus, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=4 n^{2}+2 n+1$.
From the second equation, we get
$$
[x-(2 n+1)][x-(n-2)]=0 \text {. }
$$

Solving this, we get $x_{3}=2 n+1, x_{4} \doteq n-2$.
If $x_{3}$ is an integer, then $4 n^{2}+2 n+1=2 n+1$, solving this gives $n=0$, at this time $x_{3}=1$;

If $x_{4}$ is an integer, then $4 n^{2}+2 n+1=n-2$, i.e., $4 n^{2}+n+3=0$, which has no real solutions.

In summary, when $n=0$, the sum of the squares of the two real roots of the first equation equals one integer root of the second equation.