Re 1: Not quite sure that I understand your construction but apparently it can never yield negative correlations?

Re 2: Consider the matrix $C_3=\pmatrix{1&-1/2&-1/2\cr -1/2&1&-1/2\cr -1/2&-1/2&1}$.

This is a legitimate correlation matrix (for instance of the Gaussian vector $(N,-N/2+N'\sqrt3/2,-N/2-N'\sqrt3/2)$ with $(N,N')$ i.i.d. and Gaussian) but neither $C_3$ nor any matrix whose $C_3$ is a submatrix are correlation matrices of Bernoulli vectors.

To see this, assume that $C_3$ is the correlation matrix of a random vector $X=(X_1,X_2,X_3)$ with values in $\{0,1\}^3$ and for every $i$ let $x_i^2=E(X_i)=E(X_i^2)$. Then $X_1/x_1+X_2/x_2+X_3/x_3$ is almost surely constant because the sum of the coefficients of $C_3$ is $0$. Hence $X_1/x_1+X_2/x_2$ takes exactly two values. For non degenerate $\{0,1\}$ valued random variables $X_1$ and $X_2$, this means that $x_1=x_2$. Likewise, $x_1=x_3$, hence $S=X_1+X_2+X_3$ is almost surely constant. Now $S=0$ or $S=3$ means that $X=(0,0,0)$ or that $X=(1,1,1)$, respectively, hence these cases are excluded. By the symmetry $X_i\to1-X_i$, one can assume that $S=1$ almost surely. This means that $X$ is concentrated on the three points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and, furthermore (recalling the relations $x_1=x_2=x_3$), that $X$ is uniformly distributed on these three points. Thus the correlation of $X_1$ and $X_2$ is $-1/3$ and not $-1/2$ as it should be.

Edit:

I guess the same reasoning excludes every $n\times n$ matrix $C_n$ with diagonal entries $1$ and off-diagonal entries $-1/(n-1)$, for $n\ge3$.

By the way, one sees that $C_3$ cannot be obtained through Gaussian random variables and hyperplanes as in Gowers' answer because, if it was, it would be produced by the $3\times3$ matrix with diagonal entries $\sin(1\cdot\pi/2)=1$ and off-diagonal entries $\sin((-1/2)\cdot\pi/2)=-1/\sqrt2$, which is not definite positive. (The same applies to $C_n$ for every $n\ge3$.)

Correlation matrices $C=(C_{i,j})$ of Bernoulli random vectors might be exactly those such that the matrix $(\sin(C_{i,j}\pi/2))$ is definite positive, in which case the Gaussian-cut-by-hyperplanes construction would yield them all.

2 added 43 characters in body

Re 1: Not quite sure that I understand your construction but apparently it can never yield negative correlations?

Re 2: Consider the matrix $C_3=\pmatrix{1&-1/2&-1/2\cr -1/2&1&-1/2\cr -1/2&-1/2&1}$.

This is a legitimate correlation matrix (for instance of the Gaussian vector $(N,-N/2+N'\sqrt3/2,-N/2-N'\sqrt3/2)$ with $(N,N')$ i.i.d. and Gaussian) but neither $C_3$ nor any matrix whose $C_3$ is a submatrix are correlation matrices of Bernoulli vectors.

To see this, assume that $C_3$ is the correlation matrix of a random vector $X=(X_1,X_2,X_3)$ with values in $\{0,1\}^3$ and for every $i$ let $x_i^2=E(X_i)=E(X_i^2)$. Then $X_1/x_1+X_2/x_2+X_3/x_3$ is almost surely constant because the sum of the coefficients of $C_3$ is $0$. Hence $X_1/x_1+X_2/x_2$ takes exactly two values. For non degenerate $\{0,1\}$ valued random variables $X_1$ and $X_2$, this means that $x_1=x_2$. Likewise, $x_1=x_3$, hence $S=X_1+X_2+X_3$ is almost surely constant. Now $S=0$ or $S=1$ S=3$means that$X=(0,0,0)$or that$X=(1,1,1)$X=(1,1,1)$, respectively, hence these cases are excluded. By the symmetry $X_i\to1-X_i$, one can assume that $S=1$ almost surely. This means that $X$ is concentrated on the three points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ andfurthermore, furthermore (recalling the relations $x_1=x_2=x_3$, x_1=x_2=x_3$), that$X$is uniformly distributed on these three points. Thus the correlation of$X_1$and$X_2$is$-1/3$and not$-1/2$as it should be. 1 [made Community Wiki] Re 1: Not quite sure that I understand your construction but apparently it can never yield negative correlations? Re 2: Consider the matrix$C_3=\pmatrix{1&-1/2&-1/2\cr -1/2&1&-1/2\cr -1/2&-1/2&1}$. This is a legitimate correlation matrix (for instance of the Gaussian vector$(N,-N/2+N'\sqrt3/2,-N/2-N'\sqrt3/2)$with$(N,N')$i.i.d. and Gaussian) but neither$C_3$nor any matrix whose$C_3$is a submatrix are correlation matrices of Bernoulli vectors. To see this, assume that$C_3$is the correlation matrix of a random vector$X=(X_1,X_2,X_3)$with values in $\{0,1\}^3$ and for every$i$let$x_i^2=E(X_i)=E(X_i^2)$. Then$X_1/x_1+X_2/x_2+X_3/x_3$is almost surely constant because the sum of the coefficients of$C_3$is$0$. Hence$X_1/x_1+X_2/x_2$takes exactly two values. For non degenerate $\{0,1\}$ valued random variables$X_1$and$X_2$, this means that$x_1=x_2$. Likewise,$x_1=x_3$, hence$S=X_1+X_2+X_3$is almost surely constant. Now$S=0$or$S=1$means that$X=(0,0,0)$or$X=(1,1,1)$hence these cases are excluded. By the symmetry$X_i\to1-X_i$, one can assume that$S=1$almost surely. This means that$X$is concentrated on$(1,0,0)$,$(0,1,0)$and$(0,0,1)$and furthermore, recalling the relations$x_1=x_2=x_3$, that$X$is uniformly distributed on these three points. Thus the correlation of$X_1$and$X_2$is$-1/3$and not$-1/2\$ as it should be.