439. Schoolchildren were fishing. It is known that $a_{1}$ fishermen caught at least one fish, $a_{2}$ fishermen - more than one fish, $a_{3}$ - more than two, etc., $a_{10}$ - more than nine fish. No one caught more than 10 fish. How many fish did the schoolchildren catch in total?
$\triangle$ From the condition, it follows that exactly $a_{1}-a_{2}$ fishermen caught one fish each, $a_{2}-a_{3}-$ two, $a_{3}-a_{4}-$ three, $\ldots, a_{9}-a_{10}-$ nine, $a_{10}-$ ten. Now we can calculate how many they caught together: $\left(a_{1}-a_{2}\right) \cdot 1+\left(a_{2}-a_{3}\right) \cdot 2+\left(a_{3}-a_{4}\right) \cdot 3+\ldots+\left(a_{9}-a_{10}\right) \cdot 9+a_{10} \cdot 10=a_{1}+a_{2}+a_{3}+\ldots+a_{10}$.

Answer $: a_{1}+a_{2}+\ldots+a_{10}$.