(7) How many types of triangles with integer side lengths and a perimeter of 1984 are there?
Let the three sides of the triangle be $a, b, c$ with $a \leqslant b \leqslant c$.
(i) When $a=1$, the minimum value of $c-b$ is 1, which contradicts $a>c-b$. Therefore, the minimum value of $a$ is 2. Since $a$ is the smallest side of the triangle, it must be less than (or equal to) one-third of the perimeter, so the maximum value of $a$ is 661.
(ii) From $a+b>c$, we know that the minimum value of $a+b$ is 993 (greater than half of the perimeter). Therefore, when $2 \leqslant a \leqslant 496$, after determining the value of $a$, the value of $b$ starts from $b=993-a$, and $c$ is obtained by $1984-(a+b)$, continuing until $b=c-1$ or $b=c$.
(iii) When $497 \leqslant a \leqslant 661$, it must be that $a+b>993$. After determining the value of $a$, the value of $b$ starts from $b=a$, and the value of $c$ is obtained in the same way as (ii). In summary, the list is based on the smallest side $a$ as follows:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$a$ & $b$ & $c$ & Number of types & $a$ & $b$ & $c$ & Number of types \\
\hline 2 & 991 & 991 & 1 & \multirow{2}{*}{497} & 497 & 990 & \multirow{2}{*}{247} \\
\hline 3 & 990 & 991 & 1 & & 743 & 744 & \\
\hline \multirow{2}{*}{4} & 989 & 991 & \multirow{2}{*}{2} & \multirow{2}{*}{498} & 498 & 988 & \multirow{2}{*}{246} \\
\hline & 990 & 990 & & & 743 & 743 & \\
\hline \multirow{2}{*}{5} & 988 & 991 & \multirow{2}{*}{2} & \multirow{2}{*}{499} & 499 & 986 & \multirow{2}{*}{244} \\
\hline & 989 & 990 & & & 742 & 743 & \\
\hline$\vdots$ & & & $\vdots$ & \multirow{2}{*}{500} & 500 & 984 & \multirow{2}{*}{243} \\
\hline \multirow{4}{*}{495} & 498 & 991 & \multirow{4}{*}{247} & & 472 & 742 & \\
\hline & 499 & 990 & & $\vdots$ & & & $\vdots$ \\
\hline & $\vdots$ & $\vdots$ & & are & 659 & 666 & 1 \\
\hline & 744 & 745 & & 005 & 662 & 663 & 4 \\
\hline \multirow{4}{*}{496} & 497 & 991 & \multirow{4}{*}{248} & \multirow{2}{*}{660} & 660 & 664 & \multirow{2}{*}{3} \\
\hline & 498 & 990 & & & 662 & 662 & \\
\hline & $\vdots$ & $\vdots$ & & 661 & 661 & 662 & 1 \\
\hline & 744 & 744 & & & & & \\
\hline
\end{tabular}

From Table 1, it can be seen that the number of triangles that meet the conditions can be calculated by summing four arithmetic sequences: When $a$ takes the values $2,4,6, \cdots, 496$, the corresponding number of triangles is $1,2,3, \cdots, 248$, and their sum is
$$
S_{1}=\frac{248(1+248)}{2}=30876 ;
$$

When $a$ takes the values $3,5,7, \cdots, 495$, the corresponding number of triangles is $1,2,3, \cdots, 247$, and their sum is
$$
S_{2}=\frac{247(1+247)}{2}=30628 ;
$$

When $a$ takes the values $497,499,501, \cdots, 661$, the corresponding number of triangles is 247, $244,241, \cdots, 1$, and their sum is
$$
S_{3}=\frac{83(247+1)}{2}=10292 ;
$$

When $a$ takes the values $498,500,502, \cdots, 660$, the corresponding number of triangles is 246, $243,240, \cdots, 3$, and their sum is
$$
\begin{array}{c}
S_{4}=\frac{82(246+3)}{2}=10209 . \\
S_{1}+S_{2}+S_{3}+S_{4}=30876+30628+10 \quad 292+10 \quad 209=82005 .
\end{array}
$$

Therefore, the number of triangles with integer side lengths and a perimeter of 1984 is 82005.