13. In a $2009 \times 2009$ grid, each small square is filled with a real number, the absolute value of which does not exceed 1, and the sum of the 4 real numbers in each $2 \times 2$ square is 0. Find the maximum possible sum of all the numbers in the grid.
13. We prove by induction: If the $(2 n+1) \times(2 n+1)$ number table has the property mentioned in the problem, then the sum of all numbers in the table $S_{2 n+1} \leqslant 2 n+1$. When $n=0$, it is obvious that $S_{1} \leqslant 1$. Assume $S_{2 n-1} \leqslant 2 n-1$, for the $(2 n+1) \times(2 n+1)$ number table, as shown in Figure 1, by the induction hypothesis, the sum of the numbers in the upper-left $(2 n-1) \times(2 n-1)$ part $S_{2 n-1} \leqslant 2 n-1$, and the remaining part is divided into $(n-1)+(n-1)=2(n-1)$ $2 \times 2$ number tables (the sum of the numbers in each $2 \times 2$ table is 0) and one $3 \times 3$ number table missing the upper-left corner, as shown in Figure 2. We get $S_{2 n+1}-S_{2 n-1}=a+b+c+d+e+f+g+h=(a+b+e+d)+(c+d+f+g)-d+h \leqslant 2$. Therefore, $S_{2 n+1} \leqslant S_{2 n-1}+2 \leqslant(2 n-1)+2=2 n+1$. On the other hand, if we fill each small square in the odd-numbered rows with 1 and each small square in the even-numbered rows with -1, then $S_{2 n+1}=2 n+1$. Hence, $\max S_{2 n+1}=2 n+1$, in particular, $\max S_{2009}=2009$.