a) If there are at least two non-zero numbers with the same sign in a row (WLOG positive), then argue as follows: if the middle in size number is non-negative, then the sum of the five largest numbers is greater than the largest one, contradiction; if it is negative, then the sum of the five smallest numbers is smaller than the smallest one, contradiction. Hence every row contains at most one positive and at most one negative number, hence the table overall has at least $9^2 - 9 \cdot 2 = 63$ zeroes. This is attainable, e.g. by putting $10000000(-1)$ in the first row and shifting this cyclically $8$ times, putting the results in the other rows. b) Greedily wishing for an example with as few as possible distinct numbers yields the following (with no zeroes at all): put in this order five $3$s and four $(-4)$s and shift this cyclically $8$ times, putting the results in the other rows.