= Solution
Write a test <vector> as $(u,v)$ and put $s=\sum_i u_i$, $t=\sum_jv_j$. Its <quadratic form> against the block-constant <matrix> is
$$
a(s^2+t^2)+2bst.
$$
Since $a\geq|b|$, both $a$ and $a-|b|$ are nonnegative, and
$$
a(s^2+t^2)+2bst\geq a(s^2+t^2)-2|b||st|
\geq(a-|b|)(s^2+t^2)\geq0.
$$
Equivalently it is
$$
\frac{a+b}{2}(s+t)^2+\frac{a-b}{2}(s-t)^2.
$$
Therefore the <bipartite block-constant positive semidefinite matrix> satisfies $\boxed{\begin{pmatrix}aJ_{n,n}&bJ_{n,m}\\bJ_{m,n}&aJ_{m,m}\end{pmatrix}\succeq0}$.
For positive block sizes, the condition is also necessary: choose $s=t$ and $s=-t$ to obtain $a+b\geq0$ and $a-b\geq0$. Equality is allowed, and <vectors> whose sums vanish in both blocks lie in the <kernel>. <Positive semidefiniteness>, rather than positive definiteness, is the conclusion.
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