TITLE: $10$ by $10$ board problem
QUESTION [7 upvotes]: We place the numbers $1,2,...,100$ on the $100$ squares of a $10$ by $10$ board. We then take the third greatest number from each row and create their sum $S$. Prove that at least one row has a sum of numbers that is less than $S$.

REPLY [10 votes]: Without loss of generality, we can reorder the contents of each row.  Thus, we can assume that $x_{i(j-1)}<x_{ij}$ for $1 \leq j \leq 9$.  So the third greatest entry in each cell is in the column indexed by $7$.  Without loss of generality, we can permute the rows so that $x_{(i-1)7}<x_{i7}$ whenever $1 \leq i \leq 9$.
So the matrix is:
$$\begin{pmatrix}
x_{00} & x_{01} & x_{02} & x_{03} & x_{04} & x_{05} & x_{06} & \mathbf{x_{07}} & x_{08} & x_{09} \\
x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} & \mathbf{x_{17}} & x_{18} & x_{19} \\
x_{20} & x_{21} & x_{22} & x_{23} & x_{24} & x_{25} & x_{26} & \mathbf{x_{27}} & x_{28} & x_{29} \\
x_{30} & x_{31} & x_{32} & x_{33} & x_{34} & x_{35} & x_{36} & \mathbf{x_{37}} & x_{38} & x_{39} \\
x_{40} & x_{41} & x_{42} & x_{43} & x_{44} & x_{45} & x_{46} & \mathbf{x_{47}} & x_{48} & x_{49} \\
x_{50} & x_{51} & x_{52} & x_{53} & x_{54} & x_{55} & x_{56} & \mathbf{x_{57}} & x_{58} & x_{59} \\
x_{60} & x_{61} & x_{62} & x_{63} & x_{64} & x_{65} & x_{66} & \mathbf{x_{67}} & x_{68} & x_{69} \\
x_{70} & x_{71} & x_{72} & x_{73} & x_{74} & x_{75} & x_{76} & \mathbf{x_{77}} & x_{78} & x_{79} \\
x_{80} & x_{81} & x_{82} & x_{83} & x_{84} & x_{85} & x_{86} & \mathbf{x_{87}} & x_{88} & x_{89} \\
x_{90} & x_{91} & x_{92} & x_{93} & x_{94} & x_{95} & x_{96} & \mathbf{x_{97}} & x_{98} & x_{99} \\
\end{pmatrix}$$
Let $M_i$ be the submatrix formed by the intersection of the rows $\{0,1,\ldots,i\}$ and columns $\{0,1,\ldots,7\}$.  The number $\mathbf{x_{i7}}$ is the maximum number in $M_i$.  Hence $\mathbf{x_{i7}} \geq 8(i+1)$ and so $$S \geq \sum_{i=0}^9 8(i+1)=440.$$
Let $m$ be the minimum row sum.
The sum of the entries of $M_i$ must be at most $$8(i+1) \mathbf{x_{i7}}-\left(\sum_{k=1}^{8(i+1)-1} k\right).$$  (The maximum is achieved when the numbers in $M_i$ are the $8(i+1)$ largest integers not exceeding $\mathbf{x_{i7}}$.)  Hence summing rows $0,\ldots,i$ yields $$(i+1)m \leq 8(i+1) \mathbf{x_{i7}}-\left(\sum_{k=1}^{8(i+1)-1} k\right)+100+99+\cdots+(100-2i-1).$$  This implies $m<440$ (and we are done) if any of the following are true:
\begin{align*}
\mathbf{x_{07}} \leq 33, \\
\mathbf{x_{17}} \leq 37, \\
\mathbf{x_{27}} \leq 42, \\
\mathbf{x_{37}} \leq 46, \\
\mathbf{x_{47}} \leq 50, \\
\mathbf{x_{57}} \leq 54, \\
\mathbf{x_{67}} \leq 59, \\
\mathbf{x_{77}} \leq 63, \\
\mathbf{x_{87}} \leq 67, \\
\mathbf{x_{97}} \leq 71. \\
\end{align*}
Now assume all of the above are false.  Then $$S \geq 34+38+43+47+51+55+60+64+68+72 = 532.$$  Since the average row sum is $\frac{1}{10}\sum_{i=1}^{100}i=505$, $$m \leq 505 < S$$ and we are done.