
If $x-3$ and $y+3$ are multiples of $7$, then $x$ and $y$ are congruent to $3$ and $-3$ modulo $7$, respectively.

Thus, $x^2 \equiv 3^2 \equiv 2 \pmod{7}$ and $y^2 \equiv (-3)^2 \equiv 2 \pmod{7}$.

Since $x^2+xy+y^2 \equiv 2+2+xy \equiv xy+4 \pmod{7}$, we want to find the smallest positive integer $n$ such that $xy+4 \equiv -n \pmod{7}$.

We can make a table to find the possible values of $xy$ modulo $7$:
\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
3 & -3 \\
4 & -2 \\
5 & -1 \\
6 & 0 \\
0 & 6 \\
1 & 5 \\
2 & 4 \\
\hline
\end{array}
\]
We see that $xy$ can be congruent to $-3$, $-1$, $0$, $2$, $3$, $4$, $5$ modulo $7$.

Thus, the possible values of $xy+4$ modulo $7$ are $1$, $3$, $4$, $6$, $0$, $1$, $2$.

The smallest positive integer $n$ that makes $xy+4$ a multiple of $7$ is $n=\boxed{5}$.

The answer is: 5
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Suppose $x-3$ and $y+3$ are multiples of $7$.

What is the smallest positive integer, $n,$ for which $x^2+xy+y^2+n$ is a multiple of $7$?

### Answer:
