Can a symmetric matrix always be represented as the sum of a positive-definite and negative-definite matrix?

If $X$ is symmetric then $X = (X + \lambda I) - \lambda I$. Since the eigenvalues of $X + \lambda I $ are $ \lambda_i + \lambda$ where $\lambda_i$'s are the eigenvalues of X we can find a positive $\lambda$ such that $(X + \lambda I)$ is positive definite.

Yes, see one of my questions with the details. I will type up some more:

Given $A$ such that $A = A^\top$, $A$ with both positive and negative eigenvalues, the LDU factorization will have $U=L^\top$ (follows directly from symmetry) and $D$ diagonal with both positive and negative values. So $$A=L(D_p + D_n)L^\top$$

where $D$ is separated into the positive portion $D_p$ and the negative portion $D_n$. They have all positive or all negative values and zeros. Thus when the matrix is decomposed as

\begin{align} A &= LD_pL^\top + LD_nL^\top \\ &= P + N \\ \end{align}

it is separated with $P$ symmetric positive semidefinite, and $N$ symmetric negative semidefinite.

As was pointed out in the comments $0=-1 + 1$. Thus to obtain definiteness for both, do something to $D_p + D_n$ to make it happen while retaining the value of $D = D_p + D_n$.

This answer is years late but I like this proof. The set of positive definite matrices is open in the set of symmetric matrices and must thus contain $\frac{n(n+1)}{2}$ matrices which are linearly independent. Take that as a basis for the vector space of symmetric matrices. Then every element can be written as a linear combination of positive definite matrices, and we can simply group the basis elements by the positive coefficients and negative coefficients, giving us that a symmetric matrix M can be written as the sum of a positive definite matrix and the negation of a positive definite matrix.