Answer is 438 i will ad solution later

We will prove that the answer to the problem is $438$. Let $x_i$ and $y_i$ be the number of marbles in the $i-$ row and column of the board, respectively. Note that $\displaystyle \sum_{i=1}^{10} x_i=\sum_{i=1}^{10} y_i=10$. The total sum of the numbers written on the board is $A=\sum_{i=1}^{10} (x_i(10-x_i)+y_i(10-y_i))$. We have the following key Claim: Claim: $\sum_{i=1}^{10} x_i^2 \geq 111$ and $\sum_{i=1}^{10} y_i^2 \geq 111$. Proof: We will only prove the first inequality. By Cauchy Schwarz, $\displaystyle \sum_{i=1}^{10} x_i^2 \geq \dfrac{(\sum_{i=1}^{10} x_i)^2}{10}=108.5,$ and so the sum is $\geq 109$. By $\pmod 2$ reasons, it is odd. Hence, we only need to dispose of the case it is equal to $109$. Assume that it is. Then, again by Cauchy Schwarz, $9(109-x_{10}^2)=9(x_1^2+\ldots+x_9^2) \geq (x_1+\ldots+x_9)^2=(33-x_{10})^2,$ and so we easily infer that $x_{10} \in [3,\dfrac{18}{5}]$, that is $x_10=3$. Similarly $x_i=3$ for all $i$, which is an obvious contradiction $\blacksquare$ Now, to the problem, from our Claim we have that $\displaystyle A=660-\sum_{i=1}^{10} x_i^2-\sum_{i=1}^{10} y_i^2 \leq 660-111-111=438,$ hence we have established the upper bound. Now, for the construction, denote a marble with $\times$ and an empty square with $-$. Fill in the board as follows: $\begin{bmatrix} \times & - & - & - & - & - & - & - & \times & \times \\ \times & \times & - & - & - & - & - & - & \times & - \\ \times & \times & \times & - & - & - & - & - & - & - \\ - & \times & \times & \times & - & - & - & \times & - & - \\ - & - & \times & \times & \times & - & - & - & - & - \\ - & - & - & \times & \times & \times & - & - & - & - \\ - & - & - & - & \times & \times & \times & - & - & - \\ - & - & - & - & - & \times & \times & \times & - & \times \\ - & - & - & - & - & - & \times & \times & \times & \times \\ - & - & - & - & - & - & - & \times & \times & \times \\ \end{bmatrix}$ It is easy to verify that in this case, $A=438$, and so we are done.

imo i feel like this was harder than amo p6

Answer: $438$. Example: $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline X&X&X&&&&&& \\\hline X&X&X&&&&&& \\\hline X&X&X&&&&&& \\\hline &&&X&X&X&&&\\\hline &&&X&X&X&&&\\\hline &&&X&X&X&&&\\\hline &&&&&&X&X&X&X \\\hline &&&&&&X&X&X&X\\\hline &&&&&&X&X&X&X \\\hline &&&&&&X&X&X& \\\hline \end{array} $$Proof: Construct the bipartite graph where $R_1,...,R_{10}$ are the rows and $C_1,C_2,...,C_{10}$ are the columns. There are $33$ edges thus $\sum{deg R_i}+\sum{deg C_i}=66$ Let $S$ be the sum of the numbers on the table. Every edge, whose vertices are $A$ and $B$, increases the sum $deg A(10-deg A)+deg B(10-deg B)$. \[S=\sum{degR_i(10-deg R_i)+deg C_i(10-deg C_i)}=10(\sum{deg R_i+deg C_i})-(\sum{deg R_i^2+deg C_i^2})\]\[=660-(\sum{deg R_i^2+deg C_i^2})\]Let $f=\sum{deg R_i^2}$ Claim: $f_{min}=(a_1,a_2,...,a_{10})$ has no $2$ elements whose difference is bigger than $1$. Proof: Assume that there are $2$ elements $a,a+k$. \[(a+1)^2+(a+k-1)^2=a^2+2+k^2+2k(a-1)<a^2+(a+k+1)^2\]This contradicts with our $f_{min}$ assumption since $f_{min}=f(a_1,...,a_i,...,a_j,...,a_{10})<f(a_1,...,a_i+1,...,a_j-1,...,a_{10})$ where $a_i<a_j$. Claim: $f_{min}\geq 111$. Proof: Assume that $f_{min}\leq 110$. Let elements of $f_{min}$ be $x$ times $k$ and $10-x$ times $k+1$. We have $10+10k-x=xk+(10-x)(k+1)=33\iff 10k-x=23\implies k\geq 3$. $xk^2+(10-x)(k+1)^2\leq 110$ \[110\geq xk^2+(10-x)(k^2+2k+1)=10(k+1)^2-2xk-x\]\[110\geq 10k^2+(20-2x)k+(10-x)=10k^2+(10-x)(2k+1)\geq 10k^2\implies 3\geq k\]Hence $k=3$ and $x=7$ which gives a contradiction. So \[S=660-\sum{deg R_i^2}-\sum{deg C_i^2}\leq 660-2f_{min}\leq 660-222=438\]As desired.$\blacksquare$