II. (50 points) Try to find the smallest positive integer $m$, such that the following conditions are satisfied simultaneously:
(1) $\left[\frac{2}{1977} m^{2}\right] \geqslant m+2006$ ( $[x]$ denotes the greatest integer not exceeding $x$);
(2) $99^{m}$ leaves a remainder of 11 when divided by 190.
From condition (1), we know that $m > 1977$.
Otherwise, if $m \leqslant 1977$, then we have
$$
\begin{array}{l}
{\left[\frac{2}{1977} m^{2}\right] \leqslant \frac{2}{1977} m^{2} \leqslant \frac{2}{1977} \times 1977 m} \\
= 2m \cdot 1977.
$$
Let $m = 1977 + k$ ($k$ is a positive integer), and substitute it into the inequality in condition (1):
$$
2 \times 1977 + 4k + \left[\frac{2}{1977} k^{2}\right] \geqslant 1977 + k + 2006,
$$
which simplifies to
$$
\left[\frac{2}{1977} k^{2}\right] + 3k \geqslant 29.
$$
Since $f(k) = \left[\frac{2}{1977} k^{2}\right] + 3k$ is a strictly increasing function, and $k = 10$ satisfies inequality (1), while $k = 9$ does not satisfy inequality (1), the solution to inequality (1) is $k \geqslant 10$. Therefore, the solution to the inequality in condition (1) is
$$
m = 1977 + k \geqslant 1987.
$$

Since 99 and 190 are coprime, by Euler's theorem, we have
$$
99^{1500} \equiv 1 \pmod{190},
$$
which implies $99^{72} \equiv 1 \pmod{190}$.
Lemma: If $99^a \equiv 1 \pmod{190}$, where $a$ is a positive integer, and $t$ is the smallest positive integer such that $99^t \equiv 1 \pmod{190}$, then $t \mid a$.
Proof of the lemma: Clearly, $a \geqslant t$ (by the minimality of $t$).
Let $a = t q + r$ ($q \in \mathbf{N}_{+}$, $r \in \mathbf{N}$, and $0 \leqslant r < t$).
From $99^a \equiv 1 \pmod{190}$ and $99^t \equiv 1 \pmod{190}$, we get
$$
(99^t)^q 99^r \equiv 1 \pmod{190},
$$
which simplifies to
$$
99^r \equiv 1 \pmod{190}.
$$
By the minimality of $t$ and $0 \leqslant r < t$, we must have $r = 0$. Thus, $a = t q$, which means $t \mid a$.
Returning to the original problem.
From equation (2) and the lemma, we know $t \mid 72 = 2^3 \times 3^2$.
Therefore, $t \in \{2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72\}$.
Now, we need to find the value of $t$.
Notice that
$$
99^2 \equiv 9801 \equiv 111 \pmod{190},
$$
$$
99^3 \equiv 111 \times 99 \equiv -31 \pmod{190},
$$
$$
99^4 \equiv -31 \times 99 \equiv -29 \pmod{190},
$$
$$
99^6 \equiv (-31)^2 \equiv 11 \pmod{190},
$$
$$
99^8 \equiv (-29)^2 \equiv 81 \pmod{190},
$$
$$
99^9 \equiv 11 \times (-31) \equiv 39 \pmod{190},
$$
$$
99^{12} \equiv 11^2 \equiv -69 \pmod{190},
$$
$$
99^{18} \equiv 39^2 \equiv 1 \pmod{190},
$$
so $t = 18$.
Given $99^m \equiv 11 \equiv 99^6 \pmod{190}$, and $m \geqslant 1987$, and 99 and 190 are coprime, we have $99^{m-6} \equiv 1 \pmod{190}$.
By $t = 18$ and the above lemma, we get $18 \mid (m-6)$, which means $m = 6 + 18n$ ($n \in \mathbf{N}$).
From $m \geqslant 1987$, we get $18n + 6 \geqslant 1987$. Thus,
$$
n \geqslant 111, \quad m \geqslant 18 \times 111 + 6 = 2004.
$$
Therefore, the smallest positive integer $m$ is $2004$.