16. Arrange the numbers that can be expressed as the sum of $m$ consecutive positive integers ($m>1$) in ascending order to form a sequence, denoted as $\{a(m, n)\}$, where $a(m, n)$ represents the $n$-th number (in ascending order) that can be expressed as the sum of $m$ consecutive positive integers.
（1）Write down $a(2,1), a(2,2), a(2,3), a(2,4)$;
（2）Find the analytical expression for $a(m, n)$;
（3）Can 2000 be expressed as the sum of several consecutive positive integers? If yes, find all the solutions; if not, provide the reason.
16. (1) $a(2,1)=3, a(2,2)=5, a(2,3)=7, a(2,4)=9$;
(2) Since $a(m, 1)=1+2+\cdots+m=\frac{m(m+1)}{2}$, and $\{a(m, n)\}$ is an arithmetic sequence with the first term $Q(m, 1)$ and common difference $m$, so $a(m, n)=\frac{m(m+1)}{2}+(n-1) m=m\left(n+\frac{m-1}{2}\right)$;
(3) $2000=2^{4} \cdot 5^{3}$. When $m=5$, $n+2=400$, i.e., $n=398$, so $2000=398+399+$ $400+401+402$. When $m=25$, $n+12=80$, i.e., $n=68$, so $2000=68+69+\cdots+92$. When $m=125$, $n+62=16$, i.e., $n=-46$, so there is no solution. When $m$ is even, $4000=m(2 n+$ $m-1$ ), so $2 n+m-1$ must be odd. When $m=32,2 n+m-1=125$, so $n=47,2000=$ $47+48+\cdots+78$. In summary, 2000 can be expressed as the sum of several consecutive positive integers in three ways, i.e.,
$$
2000=47+48+\cdots+78=68+69+\cdots+92=398+399+400+401+402 .
$$