$AD$ has exactly one negative eigenvalue if $v^T A v > 0$ and $D = mbox{diag}(-1,1,1)$

Fact 1: If $B$ (not necessarily symmetric) is a $3times 3$ (real) positive definite matrix, then $B^{-1}$ is also positive definite.
Proof: For non-zero $xin mathbb{R}^3$, since $B^{-1}x ne 0$, we have $x^mathsf{T}B^{-1}x = x^mathsf{T}(B^{-1})^mathsf{T}x = (B^{-1}x)^mathsf{T}B (B^{-1}x) > 0$. We are done.

Now, let $a = [1, 0, 0]^mathsf{T}$. Note that $AD = A - 2Aaa^mathsf{T}$. For $lambda < 0$, $A - lambda I$ is invertible, and begin{align} det (AD - lambda I) &= det (A - 2Aaa^mathsf{T} - lambda I) \ &= det (A - lambda I) det( I - (A - lambda I)^{-1}2Aaa^mathsf{T})\ &= det (A - lambda I) cdot left(1 - 2a^mathsf{T}(A - lambda I)^{-1}Aaright) end{align} where we have used $det (I + uv^mathsf{T}) = 1 + v^mathsf{T}u$ for real vectors $u, v$.

Let $f(lambda) = 1 - 2a^mathsf{T}(A - lambda I)^{-1}Aa$. For $lambda < 0$, by using $frac{partial Y^{-1}}{partial x} = - Y^{-1}frac{partial Y}{partial x}Y^{-1}$, we have begin{align} f'(lambda) &= - 2a^mathsf{T}(A - lambda I)^{-1}(A - lambda I)^{-1} Aa\ &= -2a^mathsf{T}[A^{-1}(A - lambda I)(A - lambda I)]^{-1}a\ &= -2a^mathsf{T}(A + lambda^2 A^{-1} - 2lambda I)^{-1}a. end{align} By Fact 1, $A^{-1}$ is positive definite. Thus, $A + lambda^2 A^{-1} - 2lambda I$ is positive definite for $lambda < 0$. By Fact 1, $(A + lambda^2 A^{-1} - 2lambda I)^{-1}$ is positive definite for $lambda < 0$. Thus, $f'(lambda) < 0$ for $lambda < 0$. Note also that $f(-infty) = 1$ and $f(0) = -1$. Thus, the equation $f(lambda) = 0$ has exactly one negative real root. As a result, $det (AD - lambda I) = 0$ has exactly one negative real root. (Q. E. D.)

Consider size $n times n$ case with $D=mbox{diag}(-1,1,dots,1)$. As @user1551 write in his answer, $AD$ has at least one negative eigenvalue.

Suppose $lambda neq eta$ is two negative eigenvalues of $AD$ with eigenvectors $v, w$, respectively; i.e. $ADv=lambda v, ADw=eta w$. Since $v$ and $w$ are linearly independent, so are $Dv$ and $Dw$. For every $s,t in mathbb R$ $sDv+tDw$ is nonzero unless $s^2+t^2=0$. It follows that $(sDv+tDw)^TA(sDv+tDw)>0$. Expand this yields $$ s^2(lambda v^T Dv) + st(lambda+eta)v^T D w+ t^2 (eta w^T D w) >0$$

Deduce that $v^TDv<0$ and $(w^TDw)(v^TDv)>(v^TDw)^2$. Define a symmetric matrix $B$ by $$B=Dvv^TD-(v^TDv)D$$

Then $Bv=0$. In other words, $v$ is an eigenvector of $B$ with eigenvalue $0$. Consider the subspace $U$ of $mathbb R^n$, given by the intersection of the orthogonal complements of subspaces generated by $v$ and $e_1=(1,0,dots,0)$; i.e. $U=langle v rangle^perp cap langle e_1 rangle ^perp$. Check that $dim U geq n-2$. For all $u in U$ we have $Bu = -(v^T D v)u $, because $v^Tu=0$ and $Du=u$. Finally, observe that $$mbox{tr}(B)=v^Tv+(n-2)(-v^T Dv) $$

This shows that $B$ is positive semi-definite. Thus $$(v^TDw)^2-(w^TDw)(v^TDv)=(w^TDv)^2-(w^TDw)(v^TDv) = w^T B w geq 0$$

contradiction.

Let us deal with the case where $A$ is $ntimes n$ for some $nge2$ and $D=operatorname{diag}(-1,1,ldots,1)$. Since $v^TAv>0$ for all nonzero $v$, every real eigenvalue of $A$ is positive. Hence $det(A)>0,,det(AD)<0$ and $AD$ has at least one negative eigenvalue. We claim that $AD$ has exactly one negative eigenvalue.

Suppose the contrary that $AD$ has at least two negative eigenvalues. By perturbing the real Jordan form of $AD$, we can pick a real matrix $B$ that is sufficiently close to $AD$, such that $B$ has at least two negative eigenvalues and is diagonalisable over $mathbb C$. Let $J=V^{-1}BV$ be the real Jordan form of $B$. Then $$ BD=VJV^{-1}D=VJleft(V^{-1}D(V^{-1})^Tright)V^T=:VJEV^T,tag{1} $$ where $E=V^{-1}D(V^{-1})^T$ is real symmetric. Let us write $$ J=pmatrix{Lambda&0\ 0&ast} text{ and } E=pmatrix{F&ast\ ast&ast}, $$ where $Lambda$ is a $2times2$ negative diagonal matrix and $F$ has the same size.

As $v^TAv>0$ for all nonzero $v$, $A$ has a positive definite symmetric part. As $B$ is close to $AD$, $BD$ is close to $A$. Hence $BD$ also has a positive definite symmetric part. Since $BD$ is congruent to $JE$ (by $(1)$) and $JE$ contains a principal submatix $Lambda F$, $Lambda F$ must have a positive definite symmetric part. It follows that all eigenvalues of $Lambda F$ have positive real parts. By matrix similarity, the eigenvalues of $(-Lambda)^{1/2}(-F)(-Lambda)^{1/2}$ have positive real parts too. But $(-Lambda)^{1/2}(-F)(-Lambda)^{1/2}$ is also real symmetric. Hence it is positive definite. So, by matrix congruence, $-F$ is positive definite and $F$ is negative definite. However, as its parent matrix $E$ has only one non-positive eigenvalue, Cauchy's interlacing inequality dictates that $F$ can have at most one non-positive eigenvalue. Hence we arrive at a contradiction and $AD$ must have exactly one negative eigenvalue at the beginning.