find the smallest positive integer so that $\det(A)\equiv 0\mod t$ 
Let $t$ be any positive integer exceeding 1. Find the smallest positive integer $k$ so that if $A$ is any $n\times n$ matrix with integer entries and with exactly $k$ entries congruent to 1 modulo t, then $\det(A)\equiv 0\mod t.$

Call an $n\times n$ matrix with integer entries an integer matrix of size n. Note that $k = n(n-1) + 2$ works. In this case, we must have at least two columns with all entries equivalent to 1 modulo t. Since there are two equal columns in $\mathbb{Z}_t,$ the determinant is zero in $\mathbb{Z}_t$.  For instance, consider $n=2, t = 2.$ In this case, we have the following matrix that shows that $k=3$ fails: $\begin{pmatrix}1 & 1\\
0 & 1\end{pmatrix}.$  For general n and t, the identity matrix shows that we cannot have $k=n.$
 A: The answer is $k = n^2 - n + 2$, and in particular it does not depend on $t$. Also, I don't see where the assumption that $t$ is odd matters at all. (Edit: See below!) Consider the $n \times n$ matrix
$$A_n = \left[ \begin{array}{cccc} 1 & 1 & 1 & \cdots \\
 0 & 1 & 1 & \cdots \\
 1 & 0 & 1 & \cdots \\
 1 & 1 & 0 & \cdots \\
 \vdots & \vdots & \vdots & \ddots \\ \end{array} \right]$$
whose first row is all $1$s and whose $k^{th}$ row has a $0$ in the $(k-1)$st entry and otherwise consists of $1$s. This matrix is invertible $\bmod t$ for any $t \ge 2$ (including $t = 2$), because after subtracting the first row from the others we get
$$\left[ \begin{array}{cccc} 1 & 1 & 1 & \cdots \\
 -1 & 0 & 0 & \cdots \\
 0 & -1 & 0 & \cdots \\
 0 & 0 & -1 & \cdots \\
 \vdots & \vdots & \vdots & \ddots \\ \end{array} \right]$$
and then after adding all the other rows to the first row we get a permutation matrix, so the determinant is $\pm 1$. $A_n$ has $n^2 - n + 1$ ones which gives $k \ge n^2 - n + 2$, and then your observation that if there are at least $n^2 - n + 2$ ones $\bmod t$ then there are at least two columns (equivalently, rows) consisting of all ones $\bmod t$ finishes it. (Again, $t$ odd is not necessary here, instead of swapping the rows just subtract one from the other.)
Edit: As user3379 indicates in the comments, this argument is incomplete and we need to construct invertible matrices with all fewer numbers of $1$s. This should be doable but possibly tedious, and it may change the answer for small values of $t$. By considering upper triangular matrices it's at least clear that $k \ge {n+1 \choose 2} + 1$.
Edit #2: The gap is not hard to fill. Start from the matrix $A_n$ above and just remove $1$s from the lower triangular portion starting from the top. After performing the same subtraction of the first row from the others we get a matrix of the form
$$\left[ \begin{array}{cccc} 1 & 1 & 1 & \cdots \\
 -1 & 0 & 0 & \cdots \\
 \ast & -1 & 0 & \cdots \\
 \ast & \ast & -1 & \cdots \\
 \vdots & \vdots & \vdots & \ddots \\ \end{array} \right]$$
where each $\ast$ is either $0$ or $-1$, and either way row reduction starting from the second row and moving down produces a matrix with determinant $\pm 1$, hence which is invertible $\bmod t$ for any $t \ge 2$. After removing all of these $1$s we can then remove $1$s above the diagonal; all of the resulting matrices are upper triangular with $1$s on the diagonal so have determinant $1$. So there exist matrices with any number of $1$s less than $n^2 - n + 2$ with determinant $\pm 1$ and hence which are invertible $\bmod t$ for any $t \ge 2$.
Edit #3: Okay, as expected there is actually now a small problem when $t = 2$ and $k < n$, because an $n \times n$ matrix with $k < n$ ones $\bmod 2$ must have the rest of its entries be zeroes, so must have a row (or column) consisting entirely of zeroes, and so can't be invertible. If $t \ge 3$ we can consider diagonal matrices with entries $\pm 1$ to deal with the $k < n$ case but this doesn't work when $t = 2$.
So for $t = 2$ the answer is $k = 1$ for $n \ge 2$ but for somewhat silly reasons; I don't know where you got this problem from but maybe that's what the restriction to odd $t$ is for.
