no body?please post your solution...

There are $29 \cdot 14$ squares below the diagonal and $29 \cdot 14$ squares above the diagonal. We have a minimum sum of $29 \cdot (1+2+\ldots+14) = 29 \cdot 105$ for the numbers below the diagonal (each number appears exactly $29$ times, so we can only have at most $29$ instances of $1$, then $29$ instances of $2$, and so on), and we have a maximum sum of $29 \cdot (29+28+\ldots+16) = 29 \cdot 315$ for the numbers above the diagonal (same reasoning). Since we reach equality, all the $15$ must lie on the main diagonal, so the central cell has $15$.