Is a "sign matrix" obtained from a symmetric positive-semidefinite matrix itself symmetic positive-semidefinite? Suppose that $A \in {\cal S}_+^n$ is a symmetric positive semidefinite matrix. Let $B = {\rm sign}(A)$, where the sign is taken elementwise. Is the resulting matrix $B$ always positive semidefinite? 
If not, under what conditions can we say that $B \in {\cal S}_+^n$ ?
 A: A small symmetric perturbation of the identity matrix is positive definite, but the corresponding sign matrix need not be positive semidefinite.  For example, let $A$ be the 3-by-3 matrix with 1 on the diagonal, and, say, -1/100 in the other 6 entries.  
(I have nothing to say about your second question.)
A: A trivial case when $B$ is semidef is say when the original matrix is elementwise non-negative (and you define sign(0) = 1). In that case, $B=ee^T$.
In general, it might be hard to give a non-trivial set of sufficient conditions on $A$ so that $B$ is semidefinite.
A: Too long for a comment.

For $2 \times 2$ real symmetric positive semidefinite matrices $A$ we have that $\text{sgn}(A)$ is symmetric positive semidefinite.

Proof.
Suppose $A := \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ is symmetric positive semidefinite with $a, b, c \in \mathbb R$.
Then $a, c \ge 0$ and $a c \ge b^2$.
Case 1.
If $b = 0$, then $\text{sign}(A)$ will be a diagonal matrix with zeros and / or ones on the diagonal and thus symmetric positive semidefinite, irrespective of whether you define $\text{sign}(0) = 0$ or $ = 1$.
Case 2.
$b \ne 0$.
In this case $a, c > 0$, because otherwise $\det(A) < 0$. Hence that the diagonal entries of $\text{sgn}(A)$ are both ones.
Since $\text{sgn}(b)^2 = 1$, the matrix $\text{sgn}(A) = \begin{pmatrix} 1 & \text{sgn}(b) \\ \text{sgn}(b) & 1 \end{pmatrix}$ is always positive semidefinite.
Remark
The same reasoning shows that (we set $\text{sgn}(0) := 0$) if a $3 \times 3$ matrix where two off-diagonal entries are zero (i.e. $\tiny\begin{pmatrix} a & 0 & 0 \\ 0 & c & f \\ 0 & f & d \end{pmatrix}$) is positive semidefinite, then so is its entrywise sign matrix.
