I like the idea of @dvitek to use pairs of multisets of partitions as a data structure for these $K_{70,70}$ decompositions. Let me repeat the idea since it partly lives in comments.

A $K_{70,70}$ decomposition is equivalent to a particular pair $\{\mathcal{A},\mathcal{B}\}$ where each of $\mathcal{A,B}$ is a multiset of $70$ partitions into distinct parts of $70.$

Assign each vertex the partition corresponding to the $i$ such that some $K_{ii}$ uses that vertex. And let $\mathcal{A,B}$ be the multisets of partitions corresponding to the two vertex classes. The following properties are satisfied:

• Among the $70$ partitions in $\mathcal{A},$ an integer $i \leq 24$ appears $i$ times and similarly for $\mathcal{B}.$

• Two partitions $\alpha,\beta$ one each from $\mathcal{A,B}$ can share at most one member. Equivalently, there is a partial edge coloring of $K_n$ using Amber and Blue so that two integers appear together in a partition $\alpha \in \mathcal{A}$ only if the corresponding edge of $K_n$ is Amber.

The converse is also true. Given such a pair of multisets of partitions, a $K_{70,70}$ decomposition is determined.

Given the second requirement, it seems that (most of) the partitions would use relatively few parts and occur to high multiplicity.

For example, perhaps $\mathcal{A}$ would use $24+23+13+10$ $10$ times and $24+17+15+14$ $14$ times ( or $a$ and $b$ times along with $24+23+14+9$ $c$ times for $a,b,c$ to be determined later subject to $a+b+c=24, a+c \leq 23, b+c \leq 9,a \leq 10, b \leq 14,c \leq 9.$) Such a start would limit the possible set of partitions using $24$ used in $\mathcal{B}$ and having chosen those, with or without their multiplicities, there might be enough restrictions to find or rule out a completion.

Alternately, there might be few enough partitions of $46$ into distinct parts (perhaps larger than $7$) to arrive at an impossibility proof.

I like the idea of @dvitek to use pairs of multisets of partitions as a data structure for these $K_{70,70}$ decompositions. Let me repeat the idea since it partly lives in comments.

A $K_{70,70}$ decomposition is equivalent to a particular pair $\{\mathcal{A},\mathcal{B}\}$ where each of $\mathcal{A,B}$ is a multiset of $70$ partitions into distinct parts of $70.$

Each particular edge belongs to a $K_{ii}$ for some $i.$ Label that edge $i.$ Assign each vertex the set consisting of the labels of its incident edges. This is a partition of $70.$ Finally, let $\mathcal{A,B}$ be the multisets of partitions corresponding to the two vertex classes. The following properties are satisfied:

• Among the $70$ partitions in $\mathcal{A},$ an integer $i \leq 24$ appears $i$ times and similarly for $\mathcal{B}.$

• Two partitions $\alpha,\beta$ one each from $\mathcal{A,B}$ can share at most one member. Equivalently, there is a partial edge coloring of $K_n$ using Amber and Blue so that two integers appear together in a partition $\alpha \in \mathcal{A}$ only if the corresponding edge of $K_n$ is Amber.

The converse is also true. Given such a pair of multisets of partitions, a $K_{70,70}$ decomposition is determined.

Given the second requirement, it seems that (most of) the partitions would use relatively few parts and occur to high multiplicity.

For example, perhaps $\mathcal{A}$ would use $24+23+13+10$ $10$ times and $24+17+15+14$ $14$ times ( or $a$ and $b$ times along with $24+23+14+9$ $c$ times for $a,b,c$ to be determined later subject to $a+b+c=24, a+c \leq 23, b+c \leq 9,a \leq 10, b \leq 14,c \leq 9.$) Such a start would limit the possible set of partitions using $24$ used in $\mathcal{B}$ and having chosen those, with or without their multiplicities, there might be enough restrictions to find or rule out a completion.

Alternately, there might be few enough partitions of $46$ into distinct parts (perhaps larger than $7$) to arrive at an impossibility proof.