Simple calculation show that:
$$ \begin{align} \det(A_2)=\begin{vmatrix} 0& 1 \\ -1& 0 \end{vmatrix}&=1\\ \det(A_4)=\begin{vmatrix} 0& 1 &1 &1 \\ -1& 0 &1&1\\ -1& -1& 0&1\\ -1& -1& -1&0 \end{vmatrix}&=1 \end{align} $$
Here is my question:
Is it true that $\det(A_{2n})=1$ for all $n\in{\mathbb Z_+}$?
With MAPLE, I tried some large $n$. And I guess it is true. But temporarily I have no idea how to show it.
Here is a combinatorial way to answer this. If we have a skew-symmetric matrix $A=\{a_{ij}\}_{1\le i,j\le 2n}$, then $\det(A)=Pf(A)^2$, where $Pf(A)$ is the Pfaffian of $A$. We know from standard methods that $$Pf(A)=\sum_{\pi \in \Pi}\text{sgn}(\pi)a_{\pi(1),\pi(2)}\cdots a_{\pi(2n-1),\pi(2n)}$$ where $\Pi$ is the set of permutations $\pi\in S_{2n}$ which satisfy $\pi(2k-1)<\pi(2k)$ for $1\le k\le n$ and $\pi(1)\le \pi(3)\le \cdots \le \pi(2n-1)$. In our case all $a_{ij}$ with $i < j$ have the same value $-1$, so we only need to prove that $$|\sum_{\pi \in \Pi}\text{sgn}(\pi)|=1.$$ To do this we will exhibit an involution on $\Pi\backslash\{id\}$ (the permutations in $\Pi$ that are not the identity).
Let $\pi \in \Pi\backslash\{id\}$, there will be a smallest $k$ so that $\pi(2k-1)= \pi(2k+1)-1$. define $\pi'$ to be the same as $\pi$ but with $\pi'(2k)=\pi(2k+2)$ and $\pi'(2k+2)=\pi(2k)$. I will leave it as an exercise for you to prove that $\pi'\in \Pi\backslash\{id\}$, $\pi''=\pi$ and that $\text{sgn}(\pi')=-\text{sgn}(\pi)$ so that $$\sum_{\pi \in \Pi}\text{sgn}(\pi)=\text{sgn}(id)=1.$$
Let $$P=\begin{pmatrix}1\\-1&1\\&-1&1\\&&\ddots&\ddots\\&&&-1&1\end{pmatrix}.$$ Then $$PA_{2n}=\begin{pmatrix}0&1&1&\ldots&1\\-1&-1\\&-1&-1\\&&\ddots&\ddots\\&&&-1&-1\end{pmatrix}.$$ Computing by row expansion, we get $\det(PA_{2n}) = 0 - (-1) + (-1) - \ldots + (-1) - (-1) = 1$. Since $\det(P)=1$, we are now done.
Edit: By considering $PA_nP^{-1}$, actually we can further show that the characteristic polynomial of $A_n$ is $p(\lambda)=\det(\lambda I_n-A_n)=\frac12\left((\lambda+1)^n+(\lambda-1)^n\right)$, regardless of whether $n$ is even or odd. Therefore, if $n$ is even and $\lambda$ is a (necessarily nonzero) eigenvalue of $A_n$, so is $1/\lambda$.
Based upon J.M.'s comments, I'd like to approach this in a different way from Davide's answer. Starting from a slightly different partitioning $$A_{2n+2} = \left(\begin{array}{cc} A_2 & B \\ -B^T & A_{2n}\end{array}\right)$$ where $B$ is a $2\times 2n$ matrix with all entries set to $1$, we know
$$\det(A_{2n+2}) = \det(A_2)\det(A_{2n} + B^T A_2^{-1} B).$$
Inspection reveals that $A_2^{-1} = A_2^T$, so
$$A_2^{-1} B = \left( \begin{array}{rrc} -1 & -1 & \ldots \\ 1 & 1 & \ldots \end{array} \right)$$
which means $B^T A_2^{-1} B = 0$. Hence, $\det(A_{2n+2}) = \det(A_{2n})\det(A_2)$. Since we know the determinant of $A_{2n}$ for $n=1\ \text{and}\ 2$ is $1$, clearly $\det(A_{2n}) = 1 \ \forall\ n \ge 1$.
Edit: it occurs to me that the inductive step is simplified by recognizing that $\det(A_{2n+2}) = \det(A_{2n})$ because $\det(A_2) = 1$. Then, this implies $\det(A_{2n}) = \det(A_2)$.
Adding a multiple of any column to another does not change the determinant of the matrix. Therefore, for each $k$ from $n$ to $2$, subtract column $k-1$ from column $k$. This leaves us with a matrix that looks like this: $$ \left[\begin{array}{r} 0&-1&0&0&\dots&0&0&0&0\\ 1&-1&-1&0&\dots&0&0&0&0\\ 1&0&-1&-1&\dots&0&0&0&0\\ 1&0&0&-1&\dots&0&0&0&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 1&0&0&0&\dots&-1&-1&0&0\\ 1&0&0&0&\dots&0&-1&-1&0\\ 1&0&0&0&\dots&0&0&-1&-1\\ 1&0&0&0&\dots&0&0&0&-1\\ \end{array}\right]\tag{1} $$ Except in column $1$, matrix $(1)$ has $-1$s on the diagonal and the superdiagonal.
Since $n$ is even, we can add the even columns to column $1$ in matrix $(1)$ and get $$ \left[\begin{array}{r} -1&-1&0&0&\dots&0&0&0&0\\ 0&-1&-1&0&\dots&0&0&0&0\\ 0&0&-1&-1&\dots&0&0&0&0\\ 0&0&0&-1&\dots&0&0&0&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&0&0&0&\dots&-1&-1&0&0\\ 0&0&0&0&\dots&0&-1&-1&0\\ 0&0&0&0&\dots&0&0&-1&-1\\ 0&0&0&0&\dots&0&0&0&-1\\ \end{array}\right]\tag{2} $$ Matrix $(2)$ has $-1$s on the diagonal and the superdiagonal. Matrix $(2)$ has determinant $(-1)^n=1$ since $n$ is even.
If $n$ were odd, then we could add the other odd columns to column $1$ in matrix $(1)$ and get $$ \left[\begin{array}{r} 0&-1&0&0&\dots&0&0&0&0\\ 0&-1&-1&0&\dots&0&0&0&0\\ 0&0&-1&-1&\dots&0&0&0&0\\ 0&0&0&-1&\dots&0&0&0&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&0&0&0&\dots&-1&-1&0&0\\ 0&0&0&0&\dots&0&-1&-1&0\\ 0&0&0&0&\dots&0&0&-1&-1\\ 0&0&0&0&\dots&0&0&0&-1\\ \end{array}\right]\tag{3} $$ Column $1$ of matrix $(3)$ is all $0$s; therefore, its determinant is $0$.
I will take the case of odd $n$ first, since my reasoning is similar but more complex in the even case.
For odd $n$, the fact that $\det A_n = 0$ follows from the fact that a skew-symmetric matrix of odd size is always singular, but here is another approach for the case in question:
Form the alternating sum of all the column vectors $v_1, v_2, \dots, v_n$ in the matrix $\sum_{i=1}^n (-1)^i v_i$.
The first row sum will be the sum $0 + 1 - 1 + 1 - \dots -1 = 0$ (an odd number of terms, one zero, which gives an even number of non-zero elements, half of them 1 and the other half -1).
An arbitrary row sum will likewise have an odd number of terms, one zero, thus an even number of non-zero terms, half of them 1 and the other half -1 (the zero is placed so that that the term before it is $(-1)^{k-1} (-1) = \pm 1$ and the element after it is $(-1)^{k+1} 1 = \mp 1$).
So $\sum_{i=1}^n (-1)^i v_i = 0$, the column vectors are linearly dependent. Therefore $\det A_n = 0$ for odd $n$.
Now for the even case.
Denote the column vectors in $A_n$ by $v_1, v_2, \dots, v_n$. For every distinct set of $n-1$ vectors $\{v_{i_1}, v_{i_2}, \dots, v_{i_{n-1}}\}$ where $i_1 < i_2 < \dots < i_{n-1}$ form an alternating sum: $$\sum_{k=1}^{n-1} (-1)^k v_{i_k}$$
Say that the vector $v_p$ is the one vector "missing" in this sum. Then, all row sums except for the $p$:th row will be 0, with the same reasoning as above; there will be an odd number of terms, exactly one will be 0, which implies an even number of non-zero terms, half of which will be 1 and the other half -1.
For the $p$:th row sum, there will be no zero terms, therefore the row sum will be either 1 or -1.
From this, it is obvious that $\det A = \pm 1$. However, as has been proven elsewhere on this site, the eigenvalues of an invertible real skew symmetric matrix comes in pairs of complex conjugates, $\lambda, \overline{\lambda}$. The determinant will therefore be $\lambda_1 \overline{\lambda_1} \cdots \lambda_{n/2} \overline{\lambda_{n/2}} = |\lambda_1|^2 \cdots |\lambda_{n/2}|^2$, which is a positive number.
We conclude that $\det A = 1$.
Let $A_{2n}$ the $2n\times 2n$ matrix defined by $(A_{2n})_{i,j} = \begin{cases} 1&\mbox{if }i<j\\ 0&\mbox{if }i=j\\ -1&\mbox{if } i>j \end{cases}$. We can show the result ($\det(A_{2n}) = 1$) by induction. The case $n=1$ and $n=2$ has been solved. Now, we can write $A_{2n+2}=\begin{pmatrix}A_{2n}&B\\-B^t &C\end{pmatrix}$ where $B$ is a $2n\times 2$ matrix whose entries are all $1$ and $C=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Consider the line $l_{2n+2}$: we can transform it, computing $l_{2n+2}\leftarrow l_{2n+2}-l_{2n+1}$. It becomes $(0,\ldots,0,-1,-1)^t$ (the $2n$ first components are $0$). After that we make $c_{2n+2}\leftarrow c_{2n+2}-c_{2n+1}$: we get $(0,\ldots,0,1,0)$. Now we conclude, expanding by the last column and after the last row (the first expansion gives $(-1)^{2n-1+2n}=-1$ and the second $(-1)^{2n-1+2n-1}\cdot (-1)$), hence we get $\det A_{2n+2}=(-1)\cdot(-1)\cdot\det A_{2n} =\det A_{2n}$.
