II. (50 points) If there exist real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying
$$
\begin{array}{l}
a_{n} \leqslant a_{n-1} \leqslant \cdots \leqslant a_{1}<2, \\
a_{1}+a_{2}+\cdots+a_{n} \geqslant n, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2},
\end{array}
$$

Find all possible values of the positive integer $n$.
Let $a_{1}, a_{2}, \cdots, a_{i}>0, a_{i+1}, a_{i+2}, \cdots, a_{n} \leqslant 0, j=n-i$. Then
$$
\begin{array}{l}
2 i>a_{1}+a_{2}+\cdots+a_{i} \\
\geqslant n-a_{i+1}-a_{i+2}-\cdots-a_{n},
\end{array}
$$

i.e., $\square$
$$
\begin{array}{l}
i-j>-a_{i+1}-a_{i+2}-\cdots-a_{n} . \\
\text { Also, }\left(-a_{i+1}-a_{i+2}-\cdots-a_{n}\right)^{2} \\
\geqslant a_{i+1}^{2}+a_{i+2}^{2}+\cdots+a_{n}^{2},
\end{array}
$$

Then $4 i>a_{1}^{2}+a_{2}^{2}+\cdots+a_{i}^{2}$
$$
\begin{array}{l}
\geqslant n^{2}-\left(a_{i+1}^{2}+a_{i+2}^{2}+\cdots+a_{n}^{2}\right) \\
\geqslant n^{2}-\left(-a_{i+1}-a_{i+2}-\cdots-a_{n}\right)^{2} \\
>(i+j)^{2}-(i-j)^{2}=4 i j .
\end{array}
$$

Since $i>0$, we have $ja_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2}$.
Thus, $n<4$.
If $n=1$, take $a_{1}=1$; if $n=2$, take $a_{1}=a_{2}=\frac{3}{2}$; if $n=3$, take $a_{1}=a_{2}=a_{3}=\frac{9}{5}$.

In conclusion, all possible values of the positive integer $n$ that satisfy the conditions are $1, 2, 3$.