5. Let the sequence $\left\{b_{n}\right\}_{n \geqslant 1}$ satisfy
$$
\begin{array}{l}
b_{200 n}=2, b_{2008}=4, \\
b_{n+3}+b_{n+2} \\
=\frac{\left(b_{n+3}+b_{n+2}\right)\left(b_{n+1}+b_{n}\right)}{2}+1(n=1,2, \cdots)
\end{array}
$$

Then $b_{1}+b_{2}+\cdots+b_{1900}=$
5.1912 .

Let $a_{n}=b_{n+1}+b_{n}$. Then
$2 a_{n+2}-2=a_{n+2} a_{n}$
$$
\begin{array}{l}
\Rightarrow a_{n+2}=\frac{-2}{a_{n}-2}(n=1,2, \cdots) \\
\Rightarrow a_{n+4}=\frac{-2}{a_{n+2}-2}=\frac{-2}{\frac{-2}{a_{n}-2}-2}=\frac{a_{n}-2}{a_{n}-1} \\
\Rightarrow a_{n+x}=\frac{a_{n+4}-2}{a_{n+4}-1}=\frac{\frac{a_{n}-2}{a_{n}-1}-2}{\frac{a_{n}-2}{a_{n}-1}-1}=a_{n} .
\end{array}
$$

Therefore, $\left\{a_{n}\right\}_{n \geqslant 1}$ is a periodic sequence with a period of 8. Then $a_{2007}=b_{2008}+b_{2007}=4+2=6$, and
$$
a_{2 \alpha 0}=a_{8 \times 200-7}=a_{7} \text {. }
$$

Thus $a_{7}=6$.
From equation (2), $\frac{a_{3}-2}{a_{3}-1}=a_{7}=6$, which gives $a_{3}=\frac{4}{5}$;
From equation (1), $\frac{-2}{a_{1}-2}=a_{3}=\frac{4}{5}$, which gives $a_{1}=-\frac{1}{2}$, and $\frac{-2}{a_{5}-2}=a_{7}=6$, which gives $a_{5}=\frac{5}{3}$.
Therefore, $a_{1}+a_{3}+a_{5}+a_{7}=\frac{239}{30}$.
Then $b_{1}+b_{2}+\cdots+b_{1920}$
$$
\begin{array}{l}
=\sum_{\substack{i=0 \\
239}}^{239}\left(b_{8 i+i}+b_{8 i+2}+\cdots+b_{8 i+8}\right) \\
=\sum_{\substack{i=0 \\
239}}\left(a_{8 i+1}+a_{8 i+3}+a_{8 i+5}+a_{8 i+7}\right) \\
=\sum_{i=0}^{1}\left(a_{1}+a_{3}+a_{5}+a_{7}\right) \\
=240 \times \frac{239}{30}=1912 .
\end{array}
$$