Is this statistical group summation unambiguously reversible? Let $X$ be a finite multisubset of $\mathbb{N}^2$. Let's introduce the following notation:
$A$ is a set of all first elements of pairs from $X$ and $B$ is a set of all second elements of pairs from $X$.
$cnt_1(n) = |\{(n,  b)|(n, b) \in X\}|$ (that is a number of elements in a corresponding set).
$cnt_2(n) = |\{(a,  n)|(a, n) \in X\}|$
$sum_1(n) = \sum_{(n,b) \in X}b$
$sum_2(n) = \sum_{(a,n) \in X}a$
$Y=\{(a, cnt_1(a), sum_1(a))|a \in A\}$
$Z=\{(b, cnt_2(b), sum_2(b))|b \in B\}$
The question: Given only $Y$ and $Z$ is it always possible to unambiguously reconstruct $X$?

Equivalently, phrased in terms of SQL: Given an (unordered) set of 2-tuples (X) of natural numbers:
CREATE TABLE X (
    a int,
    b int
);

the following statistical summaries and are derived:
Y:
SELECT a, COUNT(1) AS ca, SUM(b) AS bs FROM X GROUP BY a ORDER BY a;

Z:
SELECT b, COUNT(1) AS cb, SUM(a) AS as FROM X GROUP BY b ORDER BY b;

Is it possible to unambiguously re-construct X when knowing only Y and Z?
The answer is no . I found it helpful to think of the information in the form of a picture: mark each pair $(a, b)$ as a point on a grid, then in each column and row write both the count of the entries in that row, and the &quot;sum&quot; of the entries (the sum of the 'other' coordinate). Your example may be pictured as follows:

I'm writing the count of the dots in each row/column as the number closest the the column, and the &quot;sum&quot; as the number further away. The original data is the set of dots, the new data are Y and Z (understood to not just be sets of points, but to be positional: for example we remember that in the $Y$ data we have $(2, 3)$ in column $1$, $(1, 3)$ in column 2, and so on).
Here are two configurations of points with equal $Y$ and $Z$ data:

Hence the operation is not reversible. Per-column ambiguity arises when there are multiple integer partitions with the prescribed number of parts: eg if a column counts to 2 and sums to 5, its possible entries are 1 + 4 or 2 + 3.