. If the numbers $a_{1}, a_{2}, \ldots, a_{9}$ form a $3 \times 3$ magic square, then the numbers $a_{1}+d, a_{2}+d, \ldots, a_{9}+d$ form a $3 \times 3$ magic square, too. Hence it is sufficient to divide all the numbers into parts with equal numbers of elements: i.e. from $40 k+1$ to $40(k+1), k=0,1, \ldots, 8$. Then we need to arrange the least numbers of these parts (i.e. the numbers $1,41,81, \ldots, 321$ ) in the form of a magic square (we omit here an example, it is similar to the magic square with numbers 1 , $2, \ldots, 9)$. After that all other numbers $1+s, 41+s, \ldots, 321+s$ will also form a magic square $(s=1, \ldots, 39)$, and so do the whole sums.