Based on the comments about finding 70 partitions of 70 into distinct parts, with part $j$ appearing $j$ times among all partitions, I came up with an alternate integer linear programming formulation and found a solution.  Let $P$ be the set of all (14136) partitions of 70 into distinct parts of size at most 24.  For $j \in \{1,\dots,24\}$, let $P_j \subset P$ be the subset of partitions that contain part $j$. Let binary decision variable $x_p$ indicate whether partition $p\in P$ is used.  The problem is to find a feasible solution to the following constraints:
\begin{align}
\sum_{p\in P} x_p &= 70 \\
\sum_{p\in P_j} x_p &= j &&\text{for $j \in \{1,\dots,24\}$} \\
x_p &\in \{0,1\} && \text{for $p\in P$}
\end{align}

Here's one such solution:
```
{1,2,5,7,10,13,15,17}
{2,3,4,6,8,14,16,17}
{3,6,16,21,24}
{3,7,8,16,17,19}
{4,9,11,22,24}
{4,19,23,24}
{4,21,22,23}
{5,6,12,23,24}
{5,18,23,24}
{5,19,22,24}
{5,20,21,24}
{6,18,22,24}
{6,19,22,23}
{6,20,21,23}
{7,16,23,24}
{7,17,22,24}
{7,19,20,24}
{7,19,21,23}
{7,20,21,22}
{8,9,10,19,24}
{8,9,10,20,23}
{8,9,10,21,22}
{8,9,16,17,20}
{8,10,11,19,22}
{8,10,11,20,21}
{9,10,11,17,23}
{9,14,23,24}
{9,17,20,24}
{9,17,21,23}
{10,17,21,22}
{10,18,19,23}
{10,18,20,22}
{11,12,23,24}
{11,13,22,24}
{11,15,21,23}
{11,17,18,24}
{11,17,20,22}
{11,18,19,22}
{11,18,20,21}
{12,14,20,24}
{12,14,21,23}
{12,15,19,24}
{12,15,20,23}
{12,15,21,22}
{12,16,18,24}
{12,16,19,23}
{12,16,20,22}
{12,17,18,23}
{12,18,19,21}
{13,14,19,24}
{13,14,20,23}
{13,14,21,22}
{13,15,18,24}
{13,15,19,23}
{13,15,20,22}
{13,16,17,24}
{13,16,18,23}
{13,16,19,22}
{13,16,20,21}
{13,18,19,20}
{14,15,17,24}
{14,15,18,23}
{14,15,19,22}
{14,15,20,21}
{14,16,18,22}
{14,16,19,21}
{14,17,18,21}
{15,16,17,22}
{15,16,18,21}
{15,17,18,20}
```

**Edit**: Here's an updated formulation that captures both left ($i=1$) and right ($i=2$) sides and the rule that prevents the same pair $\{j,k\}$ from appearing together on both sides:
\begin{align}
\sum_{p\in P} x_{i,p} &= 70 &&\text{for $i\in\{1,2\}$} \\
\sum_{p\in P_j} x_{i,p} &= j &&\text{for $i\in\{1,2\}$ and $j \in \{1,\dots,24\}$} \\
x_{i,p} &\le y_{i,j,k} && \text{for $i\in\{1,2\}$ and $p\in P$ and $j<k$ with $\{j,k\}$ together in $p$} \\
y_{1,j,k} + y_{2,j,k} &\le 1 &&\text{for $1 \le j<k \le 24$} \\
x_{i,p} &\in \{0,1\} && \text{for $i\in\{1,2\}$ and $p\in P$} \\
y_{i,j,k} &\in \{0,1\} && \text{for $i\in\{1,2\}$ and $1 \le j<k \le 24$}
\end{align}

**Update**: This latest integer linear programming problem is infeasible, so the original problem has no solution.  That is, there is no way to decompose $K_{70,70}$ into 24 complete bipartite subgraphs $K_{i,i}$, $i=1,\dots,24$.