4. Find all the options for writing the number 2003 as the sum of at least 2 consecutive natural numbers.
IV /4. $2003=n+(n+1)+\cdots+m=\frac{m(m+1)}{2}-\frac{(n-1) n}{2}=\frac{m^{2}+m-n^{2}+n}{2}=\frac{(m-n)(m+n)+m+n}{2}=$ $\frac{(m-n+1)(m+n)}{2}$. Therefore, $4006=(m-n+1)(m+n)$. Since $1<m-n+1<m+n$ (the first inequality holds because the sum contains more than one number) and since 2003 is a prime number, the only possibility is $m-n+1=2$ and $m+n=2003$, so $n=1001$ and $m=1002$. The number 2003 can be written only as the sum of 2 consecutive natural numbers, i.e., $2003=1001+1002$.

Writing $2003=n+(n+1)+\cdots+m$ (or $2003=n+(n+1)+\cdots+(n+k)$): $\mathbf{1}$ point. Calculating the sum $\frac{(m-n+1)(m+n)}{2}$ (or equivalently $\mathbf{z} n$ and $k$): $\mathbf{2}$ points. Concluding that from $4006=(m-n+1)(m+n)$ it follows that $m-n+1=2$ and $m+n=2003$ (or equivalently $\mathbf{z} n$ and $k$), BECAUSE 2003 IS A PRIME NUMBER: $\mathbf{2}$ points. Completing the solution $n=1001$ and $m=1002$ (or equivalently $\mathbf{z} n$ and $k$): $\mathbf{2}$ points.

If the contestant only states the expression $2003=1001+1002$ without proving that it is the only possibility: 2 points.