jmc

Begin by factoring the two numbers. $400$ factors into $2^4\cdot 5^2$, while $576$ factors into $2^6\cdot 3^2$. In order for $400x$ to be a multiple of $576$, the prime factorization of $400x$ must include the entire prime factorization of $576$. Since the prime factorization of 576 has two more 2's and two more 3's than the prime factorization of 400, we know that the prime factorization of $x$ must include at least two 2's and at least two 3's. So, the smallest possible $x$ is $2^2\cdot 3^2=4\cdot 9=\boxed{36}$.

Checking our answer to make sure that $400\cdot (2^2\cdot 3^2)$ is a multiple of 576, we see that $$400(2^2\cdot 3^2)=2^4\cdot 5^2\cdot 2^2\cdot 3^2=5^2(2^4\cdot 2^2\cdot 3^2)=5^2(2^6\cdot 3^2)=25\cdot 576.$$