The answer is $24 \cdot 25^2 = 150\, 000$, attainable by dividing the grid into sixteen $25 \times 25$ squares and coloring each as follows: [asy][asy] pen[] p = {lightred, lightgreen, lightblue, yellow}; for (int i = 0; i < 4; i += 1) { for (int j = 0; j < 4; j += 1) { filldraw(shift(i, -j) * unitsquare, p[XOR(i, j)]); }} [/asy][/asy] Now we prove that this is maximal. Let a rainbow L be a triple of cells $(O, X, Y)$ with $O$ and $X$ in the same row, $O$ and $Y$ in the same column, and the three colors all different. Observe that each rainbow rectangle is comprised of four rainbow Ls (and these Ls are different for different rectangles), so the number of rainbow rectangles is at most one-fourth the number of rainbow Ls. Lemma. For all $0 \le x \le 100$, we have $x(100-x)^2 \le 1875(x + 50)$. Proof. The difference is $(x-25)^2 (150-x) \ge 0$. $\square$ Claim. There are at most $600\, 000$ rainbow Ls. Proof. Let $a_{i, c}$ and $b_{j, c}$ denote the number of cells of color $c \in \{1, 2, 3, 4\}$ in row $i$ and column $j$. Then the number of rainbow Ls equals \begin{align*} & \phantom{\le} \; \; \sum_{\substack{\text{cell } (i, j)\\\text{color } c}} (100 - a_{i, c})(100 - b_{j, c})\\ & \le \frac{1}{2} \sum_{\substack{\text{cell } (i, j)\\\text{color } c}} (100 - a_{i, c})^2 + (100 - b_{j, c})^2\\ & = \frac{1}{2} \sum_{i, c} a_{i, c}(100 - a_{i, c})^2 + \frac{1}{2} \sum_{j, c} b_{j, c} (100 - b_{j, c})^2\\ & \le \frac{1}{2} \sum_{i, c} 1875(a_{i, c} + 50) + \frac{1}{2} \sum_{j, c} 1875(b_{j, c} + 50)\\ & = 600\, 000. \quad \square \end{align*}It follows there are at most $\tfrac{1}{4} \cdot 600\, 000 = 150\, 000$ rainbow rectangles.

The answer is $24 \cdot 25^2 = 150\, 000$, attainable by dividing the grid into sixteen $25 \times 25$ squares and coloring each as follows: [asy][asy] pen[] p = {lightred, lightgreen, lightblue, yellow}; for (int i = 0; i < 4; i += 1) { for (int j = 0; j < 4; j += 1) { filldraw(shift(i, -j) * unitsquare, p[XOR(i, j)]); }} [/asy][/asy] Now we prove that this is maximal. Let a rainbow L be a triple of cells $(O, X, Y)$ with $O$ and $X$ in the same row, $O$ and $Y$ in the same column, and the three colors all different. Observe that each rainbow rectangle is comprised of four rainbow Ls (and these Ls are different for different rectangles), so the number of rainbow rectangles is at most one-fourth the number of rainbow Ls. Lemma. For all $0 \le x \le 100$, we have $x(100-x)^2 \le 1875(x + 50)$. Proof. The difference is $(x-25)^2 (150-x) \ge 0$. $\square$ Claim. There are at most $600\, 000$ rainbow Ls. Proof. Let $a_{i, c}$ and $b_{j, c}$ denote the number of cells of color $c \in \{1, 2, 3, 4\}$ in row $i$ and column $j$. Then the number of rainbow Ls equals \begin{align*} & \phantom{\le} \; \; \sum_{\substack{\text{cell } (i, j)\\\text{color } c}} (100 - a_{i, c})(100 - b_{j, c})\\ & \le \frac{1}{2} \sum_{\substack{\text{cell } (i, j)\\\text{color } c}} (100 - a_{i, c})^2 + (100 - b_{j, c})^2\\ & = \frac{1}{2} \sum_{i, c} a_{i, c}(100 - a_{i, c})^2 + \frac{1}{2} \sum_{j, c} b_{j, c} (100 - b_{j, c})^2\\ & \le \frac{1}{2} \sum_{i, c} 1875(a_{i, c} + 50) + \frac{1}{2} \sum_{j, c} 1875(b_{j, c} + 50)\\ & = 600\, 000. \quad \square \end{align*}It follows there are at most $\tfrac{1}{4} \cdot 600\, 000 = 150\, 000$ rainbow rectangles.

The answer is $24 \cdot 25^2 = 150\, 000$, attainable by dividing the grid into sixteen $25 \times 25$ squares and coloring each as follows: [asy][asy] pen[] p = {lightred, lightgreen, lightblue, yellow}; for (int i = 0; i < 4; i += 1) { for (int j = 0; j < 4; j += 1) { filldraw(shift(i, -j) * unitsquare, p[XOR(i, j)]); }} [/asy][/asy] Now we prove that this is maximal. Let a rainbow L be a triple of cells $(O, X, Y)$ with $O$ and $X$ in the same row, $O$ and $Y$ in the same column, and the three colors all different. Observe that each rainbow rectangle is comprised of four rainbow Ls (and these Ls are different for different rectangles), so the number of rainbow rectangles is at most one-fourth the number of rainbow Ls. Lemma. For all $0 \le x \le 100$, we have $x(100-x)^2 \le 1875(x + 50)$. Proof. The difference is $(x-25)^2 (150-x) \ge 0$. $\square$ Claim. There are at most $600\, 000$ rainbow Ls. Proof. Let $a_{i, c}$ and $b_{j, c}$ denote the number of cells of color $c \in \{1, 2, 3, 4\}$ in row $i$ and column $j$. Then the number of rainbow Ls equals \begin{align*} & \phantom{\le} \; \; \sum_{\substack{\text{cell } (i, j)\\\text{color } c}} (100 - a_{i, c})(100 - b_{j, c})\\ & \le \frac{1}{2} \sum_{\substack{\text{cell } (i, j)\\\text{color } c}} (100 - a_{i, c})^2 + (100 - b_{j, c})^2\\ & = \frac{1}{2} \sum_{i, c} a_{i, c}(100 - a_{i, c})^2 + \frac{1}{2} \sum_{j, c} b_{j, c} (100 - b_{j, c})^2\\ & \le \frac{1}{2} \sum_{i, c} 1875(a_{i, c} + 50) + \frac{1}{2} \sum_{j, c} 1875(b_{j, c} + 50)\\ & = 600\, 000. \quad \square \end{align*}It follows there are at most $\tfrac{1}{4} \cdot 600\, 000 = 150\, 000$ rainbow rectangles.