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Source. Theorem 2a, printed p. 150 (published PDF).

Statement. With finite $E,Q$, $k\ge1$ and integers $n_t\ge0$, there are pairwise disjoint partial transversals of exact sizes $n_1,\ldots,n_k$ if and only if, for every $A\subseteq E$,

$$|A|\ge\sum_{j>\sigma(E\setminus A)}n_j^* =\sum_t\bigl(n_t-\min(n_t,\sigma(E\setminus A))\bigr). \tag{1}$$

Only finitely many terms in the sum over $j$ are nonzero. The rank $\rho(E\setminus A)$ may replace $\sigma(E\setminus A)$.

Proof. Let $N=\sum_t n_t$. Any disjoint packing under the upper limits $n_t$ has union size at most $N$. It has union size $N$ exactly when each size limit is attained.

By the maximum-union formula, such a packing exists exactly when, for every $B\subseteq E$,

$$|E\setminus B|+\sum_t\min(n_t,\sigma(B))\ge N.$$

Set $B=E\setminus A$ and rearrange. This is the second expression in (1). The equality with the sum over $j$ follows by counting, for each $t$, the integers $\sigma(E\setminus A)<j\le n_t$. The rank version follows from the rank form of the same maximum formula. $\square$

No separate feasibility condition is missing: the inequalities themselves force it. In the rank version the empty-set test gives $n_t\le\rho(E)$ for every $t$, and the $A=E$ test gives $\sum_tn_t\le|E|$. Both remain valid when some or all $n_t$ are zero.