3. Let $a_{1}, a_{2}, \cdots$ be integers, for any positive integer $n$ we have
$$
a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] \text {. }
$$

If $2001 a_{199}$, find the smallest positive integer $n(n>1)$, such that $200 \mathrm{l} a_{n}$.
3. From the given, we have
$$
a_{\mathrm{T99}}=198\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right] \text {. }
$$

From $2001 a_{199}$, we get
$$
100 \left\lvert\, 99\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right]\right. \text {. }
$$

Thus, $a_{2}$ is an even number.
Let $a_{2}=2 m$. Then
$$
1001[(m-1) \times 199+2] \text {, }
$$

which simplifies to $1001(m-3)$.
Hence, let $m-3=100 t$. Then
$$
\begin{array}{l}
a_{n}=(n-1)[(100 t+2) n+2] \\
=n(n-1) \times 100 t+2\left(n^{2}-1\right) .
\end{array}
$$

Since $200 \mid a_{n}$, i.e., $100 \mid\left(n^{2}-1\right)$, $n$ must be an odd number.
Let $n=2 k+1$. Then $251 k(k+1)$.
Since $25=5^{2},(k, k+1)=1$, we have $5^{2} \mid k$ or $5^{2} \mid(k+1)$.
Thus, $k_{\min }=24, n_{\text {min }}=49$.
It is easy to verify that the smallest positive integer $n$ that satisfies the condition is
49.