Denote the numbers in the squares as shown in the figure, and let $x$ be the number of cats Ana has. If we add up both vertical arrows plus the top horizontal arrow we find that each number appears exactly once on this sum, except for $a$ which is added twice, and $e$ which does not appear on the sum. Hence, since $1+2+\ldots+9 = 45$, we have $45 - a + e = 3x$. Since $a$ and $e$ are different numbers from the set $\{1, 2, \ldots, 9\}$, we know that $-a+e$ is at least $-9+1 = -8$ and at most $-1+9 = 8$. Therefore $45 - 8 \leq 3x \leq 45 + 8 \implies 13 \leq x \leq 17$.

We will now prove that $x \neq 15$. If this were the case, then $a+b+c = e+f+g = f+h+i = 15$, but also $d+h+i = 45 - (a+b+c) - (e+f+g) = 15$. This implies $d+h+i = f+h+i$ and thus $d = f$, which is impossible.

To complete the solution, we now show with the following examples that $13$, $14$, $16$ and $17$ are possible values for $x$: