Let $a_1\leq a_2\leq \dots \leq a_n$ and $b_1\leq b_2\leq \dots \leq b_n$ be respectively the horizontal/vertical side lengths of the rectangles taken in an increasing order. By $R(a_i,b_j)$ we denote the corresponding rectangle. Let the relation $R_1\prec R_2$ means the rectangle $R_1$ can be placed inside the rectangle $R_2$. Consider the following chain of rectangles: \[R(a_1,b_1)\prec R(a_1,b_2) \prec R(a_2,b_2) \prec \dots \prec R(a_{n-1},b_{n-1})\prec R(a_{n-1},b_n)\prec R(a_n,b_n)\]There are $2n-1$ of them. One more is needed. Here comes the condition $\sum a_i = \sum b_i$. WLOG suppose $a_1\leq b_1$. Then there exists $i\,,\,1\leq i <n$ such that both $b_i$ and $b_{i+1}$ are inside the interval $[a_i,a_{i+1}]$. Indeed, otherwise we would get $\sum a_i < \sum b_i$. We've already got the missing rectangle. The chain goes like: \begin{align*} R(a_1,b_1)\prec R(a_1,b_2\prec R(a_2,b_2)\prec \dots \prec R(a_i,b_i)\prec R(a_i, b_{i+1}) & \prec \\ & \prec R(a_{i+1},b_i)\prec \\ &\prec R(a_{i+1},a_{i+1}) \prec \dots \end{align*}Now, there are $2n$ of them. That $R(a_{i+1},b_i)$ was the missing rectangle.

Let the vertical lines (n-1 of them) cut the board into n rectangles. Reorder these rectangles by their widths so that the smallest width rectangle is on the leftmost and largest width is rightmost (this will not change our answer). Similarly do this for the rows. Now we proceed with induction on n. Assume that the statement is true for $n = k$. Now, we prove for $n+1$. On the $n+1$ by $n+1$ grid, we select a $n$ by $n$ grid that is on the bottom-right. By induction, there is $2n$ rectangles that satisfy the condition. Further, we choose the upper-left rectangle and the one on the right of it. Both of these rectangles can fit into any of the other rectangles, which can be observed. Hence, we now have $2n+2$ rectangles that satisfy condition.