5. The elements of set $M$ are consecutive positive integers, and $|M| \geqslant 2, M$ the sum of the elements in $M$ is 2002, the number of such sets $M$ is $\qquad$.
5. 7 .

Let $M=\{a, a+1, a+2, \cdots, a+n-1\}$, with $n$ elements. According to the problem, $\frac{1}{2} n(a+a+n-1)=$ 2002, then $n(2 a+n-1)=4004=4 \times 7 \times 11 \times 13$. Noting that $n \geqslant 2, n$ and $2 a+n-1$ have different parities, and $n<2 a+n-1$, thus
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$n$ & 4 & $4 \cdot 7$ & $4 \cdot 11$ & $4 \cdot 13$ & 7 & 11 & 13 \\
\hline $2 a+n-1$ & $7 \cdot 11 \cdot 13$ & $11 \cdot 13$ & $7 \cdot 13$ & $7 \cdot 11$ & $4 \cdot 11 \cdot 13$ & $4 \cdot 7 \cdot 13$ & $4 \cdot 7 \cdot 11$ \\
\hline
\end{tabular}