3. Find all natural numbers $n$ such that dividing the number 2003 by $n$ leaves a remainder of 7.
IV/3. If the number 2003 gives a remainder of 7 when divided by $n$, and $n>7$, then the number $2003-7=1996$ must be divisible by $n$. From the prime factorization $1996=2^{2} \cdot 499$, we see that $n$ can be 499, $n=2 \cdot 499=998$, or $n=2^{2} \cdot 499=1996$.

Condition $n>7$: 1 point. Realization that $2003-7=1996$ is divisible by $n$: $\mathbf{2}$ points. Prime factorization $1996=2^{2} \cdot 499$: $\mathbf{1}$ point. All solutions: $\mathbf{1 + 1 + 1}$ points.