The answer is $126$. Let there be $a_i$ girls and $b_i$ boys in grade $i$, $1 \le i \le 5$. Consider a $2 \times 5$ table with $a_1, \dots, a_5$ in the first row and $b_1, \dots, b_5$ in the second row. Mark the smaller of the numbers $a_i, b_i$ for each $i$. The number of teams that can be formed is the sum of the five marked numbers. At least three marked numbers are in the same row. Suppose for instance that $a_1, a_2, a_3$ are marked. Since $b_4 \ge 19$, $b_5 \ge 19$ and $a_4+b_4 = a_5+b_5 = 100$, each of $a_4$ and $a_5$ is at most $100-19=81$. Because $a_1 + a_2 + a_3 + a_4 + a_5 = 250$, it follows that $a_1 + a_2 + a_3 = 250 - (a_4 + a_5) \ge 250 - 2 \cdot 81 = 88$.

Due to $a_i \le b_i$, $1 \le i \le 3$, the number of teams in grades 1, 2, 3 is $a_1 + a_2 + a_3$, hence it is at least $88$. Also at least $19$ teams can be formed in each of grades 4 and 5. So $88+2 \cdot 19 = 126$ teams can be formed always. The example $(a_1, a_2, a_3, a_4, a_5) = (29, 29, 30, 81, 81)$, $(b_1, b_2, b_3, b_4, b_5) = (71, 71, 70, 19, 19)$ shows that $126$ is the greatest number in question.