1. Find the smallest four-digit number $n$ for which the system

$$
\begin{aligned}
x^{3}+y^{3}+y^{2} x+x^{2} y & =n, \\
x^{2}+y^{2}+x+y & =n+1
\end{aligned}
$$

has only integer solutions.
1. Assume that the parameter $n$ is a natural number, and solve the given system in the domain of real numbers. The left side of the first equation is equal to $\left(x^{2}+y^{2}\right)(x+y)$, and thus for the numbers $s=x+y$ and $t=x^{2}+y^{2}$, we have $t \cdot s=n$ and $t+s=n+1$. The numbers $s, t$ are therefore the roots of the quadratic equation $w^{2}-(n+1) w+n=0$. From the factorization $w^{2}-(n+1) w+n=(w-n)(w-1)$, we see that $\{s, t\}=\{1, n\}$. The pair $(x, y)$ is thus a solution to the original system if and only if it is a solution to one of the systems

$$
\left\{\begin{array} { r l } 
{ x + y } & { = 1 , } \\
{ x ^ { 2 } + y ^ { 2 } } & { = n }
\end{array} \quad \text { or } \quad \left\{\begin{array}{rl}
x+y & =n \\
x^{2}+y^{2} & =1
\end{array}\right.\right.
$$

In the first case, the numbers $x, y$ are the roots of the quadratic equation $z^{2}+(1-z)^{2}=n$. Solving this, we get

$$
\{x, y\}=\left\{\frac{1+\sqrt{2 n-1}}{2}, \frac{1-\sqrt{2 n-1}}{2}\right\}
$$

Similarly, from the second system in (1), it follows that the numbers $x, y$ are the roots of the quadratic equation $z^{2}+(n-z)^{2}=1$. Its discriminant $D=4\left(2-n^{2}\right)$ is non-negative only for $n=1$ (recall that $n$ is a natural number), in which case both systems in (1) are identical, so no other solutions exist besides (2).

We have found that for every natural number $n$, all solutions to the original system in the domain of real numbers are described by the relation (2). These are integers if and only if the value $2 n-1$ is a square (of an odd natural number). If the number $n$ is a four-digit number, then $n \geqq 1000$, and thus $2 n-1 \geqq 1999$. Since $44^{2}=1936$ and $45^{2}=2025$, the desired number $n$ is determined from the equation $2 n-1=2025$. Hence, $n=1013$.

Note that in the first part of the solution, we can also proceed as follows: in the first equation $\left(x^{2}+y^{2}\right)(x+y)=n$, we can substitute the expression $x^{2}+y^{2}=n+1-(x+y)$ from the second equation for the first factor. This gives us a quadratic equation for the unknown $s=x+y$: $(n+1-s) s=n$, whose roots are $s_{1}=1$ and $s_{2}=n$.