Solution:

Note that rearranging the columns does not change the medians, hence we may sort the first row, so that $d_{1,1} \leq d_{1,2} \leq \ldots \leq d_{1,7}$. The calculations are much simplified if we subtract $i-1$ from each row. In other words, we put $D_{i, j}=d_{i, j}-(i-1)$. This subtracts $i-1$ from the median $m_{i}$ as well - that is if $M_{i}$ is the median of $D_{i, j}$'s, then $M_{i}=m_{i}-(i-1)$. Thus the sum of the $M_{i}$'s is equal to the sum of the $m_{i}$'s minus $0+1+2+\ldots+9=45$. We shall show that sum of $M_{i}$'s is 0, so that the sum of the $m_{i}$'s is 45 and the average is always 4.5.

Note that since $D_{1,1} \leq D_{1,2} \leq \ldots \leq D_{1,7}$ the entry $D_{1,4}$ is a median. The fourth column will continue to contain a median until $d_{i, 7}=0$ (at which point the third column will contain a median), that is $10-D_{1,7}$ times (note that $d_{1,7}=D_{1,7}$). The sum of those medians is then equal $D_{1,4}(10-D_{1,7})$. After that, median moves to the third column and stays there until $d_{i, 6}=0$ (this may be no time at all, if $d_{1,6}=d_{1,7}$, but that will not affect the calculation). The contribution of those medians is $D_{1,3}(D_{1,7}-D_{1,6})$. Continuing this way we see that the medians in the second column contribute $D_{1,2}(D_{1,6}-D_{1,5})$ and ones in the first column $D_{1,1}(D_{1,5}-D_{1,4})$. A median then moves to the seventh column, but by that point its value has dropped, $D_{i, 7}=D_{1,7}-10$. The contribution of those medians is then $(D_{1,7}-10)(D_{1,4}-D_{1,3})$. Similarly for those in sixth and fifth columns we get $(D_{1,6}-10)(D_{1,3}-D_{1,2})$ and $(D_{1,5}-10)(D_{1,2}-D_{1,1})$. Finally the median moves to the fourth column again, staying there remaining $D_{1,1}$ times, contributing $(D_{1,4}-10) D_{1,1}$. Overall, the sum of all medians is thus
$$\begin{array}{r} D_{1,4}(10-D_{1,7})+D_{1,3}(D_{1,7}-D_{1,6})+D_{1,2}(D_{1,6}-D_{1,5})+ \\ D_{1,1}(D_{1,5}-D_{1,4})+(D_{1,7}-10)(D_{1,4}-D_{1,3})+(D_{1,6}-10)(D_{1,3}-D_{1,2}) \\ +(D_{1,5}-10)(D_{1,2}-D_{1,1})+(D_{1,4}-10) D_{1,1} . \end{array}$$
It is fairly easy to see that this expression is in fact equal to 0 (for example, by considering the linear and quadratic terms separately). This means that the sum of new medians $M_{i}$ is zero, and the sum of the original $m_{i}$'s is 45, as wanted.

Solution:

We will prove a stronger claim: for all $a$, the number of $m_{i}$'s equal to $a$ equals the number of $m_{i}$'s equal to $9-a$. (By a pairing argument, this implies that the average of the $m_{i}$'s is $9 / 2$.) Indeed, for $1 \leq j \leq 10$ let $M_{i, j}$ denote the $j$th smallest entry in row $i$ of the table (so that $m_{i}=M_{i, 4}$); we will show that for all $a$ and $j$, the number of $M_{i, j}$'s equal to $a$ equals the number of $M_{i, 8-j}$'s equal to $9-a$.

Henceforth, all row-indices are to be interpreted modulo 10, and "between" is meant in the inclusive sense.

It follows from the defining property of the table that for all $i$ between 1 and 10, all $a$ between 0 and 9, and all $k$ between 0 and $a$, the number of $k$'s in row $i$ equals the number of $k+a$'s in row $i+a$. Replacing $a$ by $9-a$, and summing over all $k$ between 0 and $a$, we find that the number of entries between 0 and $a$ in row $i$ equals the number of entries between $9-a$ and 9 in row $i+9-a$. Hence for all $j$, the number of entries between 0 and $a$ in row $i$ is greater than or equal to $j$ if and only if the number of entries between $9-a$ and 9 in row $i+9-a$ is greater than or equal to $j$. But this means that the $j$ smallest entries in row $i$ are all between 0 and $a$ if and only if the $j$ largest entries in row $i+9-a$ are all between $9-a$ and 9. That is, $M_{i, j} \leq a$ if and only if $M_{i+9-a, 8-j} \geq 9-a$. Replacing $a$ by $a-1$, we see also that $M_{i, j} \leq a-1$ if and only if $M_{i+10-a, 8-j} \geq 10-a$. Combining the last two facts, we conclude that $M_{i, j}=a$ if and only if $M_{i+10-a, 8-j}=9-a$. Summing over $i$ (and noting that $i+10-a$ varies over $0,1, \ldots, 9 \bmod 10$ as $i$ does), we see that the number of $i$'s with $M_{i, j}=a$ equals the number of $i$'s with $M_{i, 8-j}=9-a$, as was claimed above.