Example 5 Find the smallest positive integer $n$, such that any $n$ different positive integers taken from the set $I=\{1,2, \cdots, 999\}$ contain 4 different numbers $a, b, c, d$ satisfying $a+2 b+3 c=d$.
Consider the set $S=\{x, x+1, x+2, \cdots, 999\}\left(x \in \mathbf{N}_{+}\right)$. For any 3 distinct numbers $a, b, c$ in $S$, we have $a+2 b+3 c \geqslant(x+2)+2(x+1)+3 x=6 x+4$. If $6 x+4 \geqslant 1000$, i.e., $x \geqslant 166$, then there does not exist $d$ in $S$ such that $a+2 b+3 c=d$. Therefore, $S=\{166,167,168, \cdots, 999\}$ does not contain 4 distinct positive integers $a, b, c, d$ satisfying $a+2 b+3 c=d$. And $|S|=999-165=834$. Hence, the smallest positive integer $n \geqslant 835$.

When $n \geqslant 835$, for any $n$-element subset $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1}a_{1}$, and $a_{n}-3 a_{i}-2 a_{1} \neq a_{i}$, then $a_{n}-3 a_{i}-2 a_{1} \notin S$ (otherwise, $a_{n}-3 a_{i}-2 a_{1}=a_{j} \in S, j \neq 1, i, n$. Thus, $a_{j}+2 a_{1}+3 a_{i}=a_{n}$, which contradicts the assumption).
Take $i_{0}=169-a_{1}\left(i_{0}\right.$ can be determined by the last equality in (*)).
Then
\[
\begin{aligned}
\left(a_{n}-3 a_{i_{0}}-2 a_{1}\right)-a_{1} & =a_{n}-3\left(a_{i_{0}}+a_{1}\right) \geqslant a_{n}-3\left[a_{n}-\left(n-i_{0}\right)+a_{1}\right] \\
& =3 n-2 a_{n}-3\left(a_{1}+i_{0}\right)=3 \times 835-2 \times 999-3\left(a_{1}+i_{0}\right) \\
& =3\left[169-\left(a_{1}+i_{0}\right)\right]=0 .
\end{aligned}
\]

Thus, for $k=2,3,4, \cdots, 168-a_{1}\left(a_{n}-3 a_{i_{0}}-2 a_{1}\right)-a_{1} \geqslant 0$.
\[
\text{That is, } a_{n}-3 a_{k}-2 a_{1}>a_{1}, \text{ and } a_{n}-3 a_{k}-2 a_{1}=a_{k} \text{ has at most one solution } a_{k}=\frac{1}{4}\left(a_{n}-2 a_{1}\right).
\]
Hence, for $k=2,3, \cdots, 168-a_{1}$, $a_{n}-3 a_{k}-2 a_{1}$ will produce $166-a_{1}$ numbers greater than $a_{1}$ and not equal to $a_{k}$, i.e., at least $166-a_{1}$ numbers in $\left\{a_{1}, a_{1+1}, \cdots, 999\right\}$ do not belong to $S$. Therefore, there are at least $166-a_{1}$ numbers greater than $a_{1}$ in $I$ that do not belong to $S$. And the $a_{1}-1$ numbers less than $a_{1}$ also do not belong to $S$. Hence, there are at least $a_{1}-1+166-a_{1}=165$ numbers in $I$ that do not belong to $S$. Thus, the number of elements in $S$ is at most $999-165=834$. This contradicts $n=|S|=835$. Therefore, there exist 4 distinct numbers $a, b, c, d \in S$ such that $a+2 b+3 c=d$.
In summary, the smallest value of $n$ is 835.
(2) Special Case Estimation
Examine some special cases that satisfy the conditions to estimate the maximum (minimum) value.

---

The translation preserves the original text's formatting and structure.