For example, 32006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find
$$
a_{1}+a_{2}+\cdots+a_{2000}
$$

the minimum value. ${ }^{[1]}$
First, we prove: For any 119 positive integers \( b_{1}, b_{2}, \cdots, b_{119} \), there must exist a subset (at least one, or all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
\[ b_{1}, b_{1}+b_{2}, b_{1}+b_{2}+b_{3}, \cdots, b_{1}+b_{2}+\cdots+b_{119}. \]
If one of them is a multiple of 119, then the conclusion holds.

If none of them is a multiple of 119, then the remainders when they are divided by 119 can only be \( 1, 2, \cdots, 118 \), which are 118 different cases.

By the pigeonhole principle, we know that there must be two of them that have the same remainder when divided by 119. Let's assume these two are
\[ b_{1}+b_{2}+\cdots+b_{i} \]
and \( b_{1}+b_{2}+\cdots+b_{j} (1 \leqslant i < j \leqslant 119) \).
Then, \( 119 \mid \left(b_{i+1}+b_{i+2}+\cdots+b_{j}\right) \).

Next, for any 119 numbers among \( a_{1}, a_{2}, \cdots, a_{2006} \), by the previous proof, we know that there must be a subset of these numbers whose sum (let's denote it as \( x \)) is a multiple of 119.

By the problem's condition, we know that \( x \) is not equal to 119, so \( x \) is at least \( 2 \times 119 \).
Since \( 2006 = 16 \times 119 + 102 \), we have
\[ \begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2006} \\
\geqslant 16 \times 238 + 102 = 3910.
\end{array} \]

If we set \( a_{119}=a_{238}=\cdots=a_{1904}=120 \), and the rest of the numbers are 1, then the equality holds.

Therefore, the minimum value of \( a_{1}+a_{2}+\cdots+a_{2006} \) is 3910.