$\qquad$ Arrange 1999 positive integers in a row so that the sum of any consecutive $n$ terms $(n=1,2, \cdots$, 1999 ) is not 119. Find the minimum value of the sum of these 1999 numbers.
Parse: From left to right, let these 1999 numbers be $a_{1}, a_{2}, \cdots, a_{1999}$, and set $b_{i}=a_{1}+a_{2}+\cdots+a_{i}(i=1,2, \cdots, 1999)$, we will prove that the sum of any 119 consecutive terms is not less than 238. Without loss of generality, we prove that $b_{119} \geqslant 238$.

Proof by contradiction. If $b_{119}b_{j}-b_{i} \geqslant 2 \times 119$, it leads to a contradiction! Therefore, $b_{119} \geqslant 238$, and similarly, we can prove that the sum of any 119 consecutive terms is not less than 238.
$1999=119 \times 16+95$, thus $b_{1999} \geqslant 238 \times 16+95=3903$, the equality can be achieved in the following scenario: