In the $n^\text{th}$ row, there are $2n$ numbers, and the difference between consecutive terms is $2$. Now, we want to find a function $f(n)$ that will generate the first term of the $n^\text{th}$ row. Since the difference between consecutive first terms is $3$, then $7$, then $11$, and so on, and the difference between consecutive terms in this sequence is $4$, a constant, $f(n)$ is quadratic. Now, letting $f(n)=an^2+bn+c$, we have $a+b+c=2$, $4a+2b+c=5$, and $9a+3b+c=12$. Subtracting the first equation from the second, and multiplying both sides by $3$, we find $9a+3b=9$. Plugging into the third equation, $9+c=12 \implies c=3$. Thus, $a+b+3=2 \implies a+b=-1$ and $4a+2b+3=5 \implies 2a+b=1$. Subtracting, we have $2a+b-a-b=1-(-1)\implies a=2 \implies b=-3$. Thus, $f(n)=2n^2-3n+3$. Plugging in $n=10$, we find that the first number in tenth row is $2\cdot 10^2 -3\cdot10+3=173$. Therefore, the tenth row is $173,175,177,\cdots,209,211$. The sum of these numbers is $\boxed{3840}$.

So, we want to find the first element of the $n^{\text{th}}$ row. The first number is $2$. Adding $3$ gives the next number, adding $7$ the next, adding $11$ the next, and so on. Hence, the first element of the $n^{\text{th}}$ row is $$2+(4*1-1)+(4*2-1)+...(4*(n-1)-1)=2-(n-1)+4(1+2+...(n-1))$$$$=3+2n(n-1)-n=2n^2-3n+3.$$Once we get the first element of the $n^{\text{th}}$ row, there will be $2n$ elements in the row total, each one is $+2$ the one preceding. Hence, we sum to get $$2n\left(2n^2-3n+3\right)+2(1+2+...(2n-1))=2n(2n^2-n+2).$$Plug in $n=10$ to get $3840.$