WakeUp wrote: $162$ pluses and $144$ minuses are placed in a $30\times 30$ table in such a way that each row and each column contains at most $17$ signs. (No cell contains more than one sign.) For every plus we count the number of minuses in its row and for every minus we count the number of pluses in its column. Find the maximum of the sum of these numbers. $162+144=306,$ isn't it?

Let's call this sum $S$. Clearly $S\leq 306$. Note that if we put a plus sign on each row, then we count ALL the minus signs. Likewise, if we put a minus sign on each column, we count all the plus signs. Put $30$ pluses along one of the main diagonals and $30$ minuses along the other (a main diagonal is a diagonal that joins two opposite corners). Since $30$ is even, these two diagonals do not intersect. Fill in the rest of the signs such that each row and column has at most $17$ signs, clearly it is possible since $17\times 30=510>306$. Then each row has a plus sign, and each column has a minus sign, so all minuses and pluses are counted in $S$ and $S=306$. A related question, can S take any integer value between 0 and 306?

Sorry, I made a typo on the previous post. I meant $S\geq 306$. I believe the maximum value of $S$ is $S=2592$.Consider the following arrangement: Note that $306=17\times 18$, consider an $18\times 18$ square with the top left-bottom right diagonal empty. Arrange the $162$ pluses in two $9\times 9$ squares and put them on the top right & bottom left corners. Fill the rest of the squares with minuses. Note that in this arrangement: -Every row & column has exactly 17 signs. -Every row & column has 9 pluses and 8 minuses. Therefore: Every minus counts $9$ pluses. In total $144\times 9=1296$ Every plus counts $8$ minuses. In total $162\times 8=1296$ So in total $S=2592$. However I have no rigorous proof of why this is the maximum. Any help appreciated!

A related question, can S take any integer value between 0 and 2592?