The answer is $n$ odd. Let $I$ denote the $n\times n$ identity matrix, and let $A$ denote the $n\times n$ matrix all of whose entries are $1$. If $n$ is odd, then the matrix $A-I$ satisfies the conditions of the problem: the dot product of any row with itself is $n-1$, and the dot product of any two distinct rows is $n-2$. Conversely, suppose $n$ is even, and suppose that the matrix $M$ satisfied the conditions of the problem. Consider all matrices and vectors mod $2$. Since the dot product of a row with itself is equal mod $2$ to the sum of the entries of the row, we have $M v = 0$ where $v$ is the vector $(1,1,\ldots,1)$, and so $M$ is singular. On the other hand, $M M^T = A-I$; since $$(A-I)^2 = A^2-2A+I = (n-2)A+I = I,$$ we have $(\det M)^2 = \det(A-I) = 1$ and $\det M = 1$, contradicting the fact that $M$ is singular.