6. Given that for every real number $x$ and $y$, the function $f(x)$ satisfies $f(x)+f(y)=f(x+y)+x y$. If $f(1)=m$, then the number of positive integer pairs $(m, n)$ that satisfy $f(n)=1998$ is $\qquad$.
6.16.

Let $y=1$, we get
$$
\begin{array}{l}
f(x)+f(1)=f(x+1)+x, \\
\therefore f(x+1)-f(x)=f(1)-x, \\
\therefore f(x)-f(x-1)=f(1)-(x-1), \\
\quad f(x-1)-f(x-2)=f(1)-(x-2), \\
\quad \cdots \cdots . \\
\quad f(2)-f(1)=f(1)-1 .
\end{array}
$$

Adding the above $x-1$ equations, we get
$$
\begin{array}{l}
f(x)-f(1)=(x-1) f(1)-\frac{1}{2} x(x-1) \text {. } \\
\therefore f(x)=x f(1)-\frac{1}{2} x(x-1) \\
=x m-\frac{1}{2} x(x-1) \text {. } \\
\because f(n)=1998 \text {, } \\
\therefore 1998=m n-\frac{1}{2} n(n-1) \text {. } \\
\therefore n(2 m-n+1)=4 \times 3^{3} \times 37 \text {. } \\
\because n+(2 m-n+1)=2 m+1 \text { is odd. } \\
\end{array}
$$
$\therefore n$ and $2 m-n+1$ are one odd and one even.
When $n$ is even, the number of odd values for $2 m-n+1$ is $(3+1)(1+1)=8$ (values), i.e., there are 8 solutions.
When $n$ is odd, $n$ can take 8 odd values, also giving 8 solutions. Therefore, the number of positive integer pairs $(m, n)$ is 16.