4. We will call a number anti-triangular if it can be written in the form $\frac{2}{n(n+1)}$ for some natural number $n$. For how many numbers $k(1000 \leqslant k \leqslant 2000)$ can the number 1 be written as the sum of $k$ anti-triangular numbers (not necessarily distinct)?
Answer: all 1001 possible values of $k$ work.

Notice that $1 / 3=2 / 2 \cdot 3$ and $1 / 6=2 / 3 \cdot 4$ are anti-triangular numbers. Therefore, unity can be written as the sum of three, four, and five such numbers: $1=1 / 3+1 / 3+1 / 3=1 / 6+1 / 6+1 / 3+$ $1 / 3=1 / 6+1 / 6+1 / 6+1 / 6+1 / 3$. We will prove that the number of addends in the representation can be increased by 3 - from this it will follow that all $k \geqslant 3$ work. Indeed,

\[
\begin{aligned}
\frac{2}{n(n+1)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right) & =4\left(\frac{1}{2 n}-\frac{1}{2 n+2}\right)=4\left(\frac{1}{2 n}-\frac{1}{2 n+1}+\frac{1}{2 n+1} \frac{1}{2 n+2}\right)= \\
=4\left(\frac{1}{2 n(2 n+1)}+\frac{1}{(2 n+1)(2 n+2)}\right) & =\frac{2}{2 n(2 n+1)}+\frac{2}{2 n(2 n+1)}+\frac{2}{(2 n+1)(2 n+2)}+\frac{2}{(2 n+1)(2 n+2)}
\end{aligned}
\]

i.e., one anti-triangular number can be replaced by the sum of four others, thereby increasing the number of addends by three.