Edit: nvm this is wrong - see below

laegolas wrote: Thus the table must have exactly $7$ distinct colors. Your argument doesn't work after the first row/column. In fact, it's possible to use way more than 7 colours (I don't know if my example is the maximun or not)

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If each row must have distinct number of colors, we have exactly one row with one color, one row with two colors and so on to seven. The same with columns. So let paint one row with one color ( it can be the uppermost ). Next row we can paint with two colors but if they are different from the first one, we would not get a column with exactly one color, so we can get at most one new color. Let consider the row with three colors. If they are different from the first color we would not get a column with exactly one color which is necessary. So we get at most two new colors and so on for four, five, six and seven. So we would get at most $1 + \frac{7(0 + 6)}{2} = 22$ colors. $$\left[ \begin{array}{ccccccc} A & A & A & A & A & A & A\\ A & A & A & A & A & A & B\\ A & A & A & A & A & C & D\\ A & A & A & A & E & F & G\\ A & A & A & H & I & J & K\\ A & A & L & M & N & O & P\\ A & Q & R & S & T & U & V\\ \end{array} \right]$$