Let $A$ be an $n \times n$ symmetric integer matrix with even entries on the main diagonal. Then (i) if $n \equiv 0$ mod $4$, then Let $A$ be an $n \times n$ symmetric integer matrix with even entries on the main diagonal.
Then (i) if $n \equiv 0$ mod $4$, then det$(A)\equiv $ 0 or 1, mod 4; (ii) if $n \equiv 2 $mod $4$; then det$(A) \equiv 0$ or $−1$ mod $4$.
In above, by even entries on the main diagonal, I mean that the entries are lying in $\{0,\pm2,\pm4,\pm6,\ldots\}$
 A: Let $n=2N.$ For $N=1$ by inspection we get
$$\det A=0\  {\rm or}\ \det A= -1 \mod (4) \quad (*)$$
Every element $a_{ij}$ can be expressed as
$$a_{ij}=b_{ij}+4d_{ij},\qquad d_{ij}=d_{ji}\in \mathbb{Z},\ b_{ij}=b_{ji}\in \{\pm 2, \pm 1,0\}$$
As the determinant is linear with respect to each row we get
$$\det A=\det B\ \mod(4)$$ and the diagonal entries of $B$ are even numbers.
Observations

*

*the determinant of $B$ does not change if the integer multiple of $k$th row is subtracted from $l$th row, $k\neq l$

*the determinant does not change if the same multiple of the $k$th column is subtracted from the $l$th column

*changing the indices $i\longleftrightarrow j,$ i.e. exchanging the $i$th and the $j$th columns and rows does not change the determinant

*the new matrix is still symmetric and has even entries on the main diagonal

If one of the rows of $B$ is  a zero row, then $\det B=0.$ So we may restrict to the case, when every row (by symmetry  every column) of $B$ contains nonzero entries.
For $k$th row choose an entry with the least nonzero absolute value in this row, say $b_{km}$ (it may occur that $k=m$).
Consider a nondiagonal entry $b_{kl},$ $l\neq m$ and $l\neq k.$  Then $b_{kl}=cb_{km}$ for an integer number $c.$ Subtract $c$ times the $m$th column from the $l$th column and at the same time subtract $c$ times the $m$th row  from the $l$th row. In this way we obtain a new matrix $B'$ with the same determinant and with the same properties as $B,$ according to Observations. Moreover $b'_{kl}=b'_{lk}=0.$ We can keep repeating that procedure until we obtain a matrix $C$ such that every row (hence every column) contains at most one nonzero off diagonal entry.
In what follows we will modify the matrix $C$ exchanging the rows and columns, but we will keep the same symbol $C$ for all matrices on the way.
If $C$ is diagonal, then $\det C=0 \mod(4).$ If not, consider the first row (say $k$th row) of $C$ containing a nonzero off diagonal entry.
We can change the indices
$1\longleftrightarrow k$ so the first row of $C$
contains a nonzero off diagonal entry $c_{1m},$ $m>1.$
Exchange the indices $2\longleftrightarrow m.$ In this way the  matrix $C$ has the property $c_{12}=c_{21}\neq 0.$ As every row may contain at most one nonzero off diagonal entry, we get $c_{1l}=c_{2l}=0$ for $l\ge 3.$ By symmetry we obtain $c_{l1}=c_{l2}=0$ for $l\ge 3.$
In this way the matrix $C$ is a direct sum of $2\times 2$ matrix $$\begin{pmatrix} c_{11} & c_{12}\\
c_{12} & c_{22}
\end{pmatrix},\quad c_{12}\neq 0\qquad (**)$$ and $(n-2)\times (n-2)$ symmetric matrix $D$ with even diagonal entries. We can  apply the above procedure to $D$ and to  next matrices we obtain on the way. Eventually the matrix $C$ will be represented as a direct sum of $2\times 2$ matrices and a diagonal matrix $E$ of even order (as $n=2m$ is even) with even elements on the diagonal.  The determinant of $C$ is the product of the determinants of those matrices. If $E\neq 0,$ then $\det C=0 \mod(4).$ If $E=0$ we obtain the product of $N=n/2$ determinants of $2\times 2$ matrices of the form $(**).$  If the determinant of one of those $2\times 2$ matrices is divisible by $4,$ then $\det C=0\, \mod(4).$ Otherwise all the determinants of those $2\times 2$ matrices are equal $-1\mod 4,$ hence the product is equal $(-1)^N\mod(4),$ and the conclusion follows.
A: This proof is inspired by the basic facts about Pfaffians. Recall that, if $A$ is a $(2m) \times (2m)$ skew symmetric matrix, then $\det(A) = \text{Pf}(A)^2$, where the Pfaffian, $\textbf{Pf}(A)$, is a certain sum over the perfect matchings of the complete graph $K_{2n}$. In our situation, we have a symmetric matrix, but we'll prove something similar modulo $4$.
Define a "perfect matching" to be a partition of $\{ 1,2,\ldots, 2m \}$ into $m$ disjoint two element sets. For example, $\{ 1,2,3,4 \}$ has three perfect matchings: $\{1,2 \} \cup \{ 3,4 \}$, $\{1,3 \} \cup \{ 2,4 \}$ and $\{1,4 \} \cup \{ 2,3 \}$. Let $A$ be a $(2m) \times (2m)$ symmetric matrix. The Hafnian of $A$ is defined by
$$\text{haf}(A) = \sum_M \prod_{\{ i,j \} \in M} A_{ij}$$
where the sum is over all perfect matchings. For example, the hafnian of a $4 \times 4$ symmetric matrix is $A_{12} A_{34} + A_{13} A_{24} + A_{14} A_{23}$.
I will show that
$$\det(A) \equiv (-1)^m \text{haf}(A)^2 \bmod 4. \qquad (\ast)$$
Proof: Expand both sides of $(\ast)$, and group like terms together, using that $A_{ij} = A_{ji}$. Each term is a product of $m$ $A_{ij}$'s. Think of each term as $m$ edges in the complete graph $K_{2m}$; together they form a subgraph of $K_{2m}$ (possibly with loops or repeated edges) in which each
vertex has degree $2$. Thus, each component of this graph is a cycle of some length; let the lengths of the cycles be $a_1$, $a_2$, ..., $a_r$ with $\sum a_i = 2m$. We emphasize that this is an undirected graph: For example, the terms $A_{12} A_{23} A_{34} A_{41}$ and $A_{14} A_{21} A_{32} A_{43}$ have been combined.
On the left hand side of $(\ast)$, the coefficient of our monomial is $\pm 2^{\# \{i : a_i \geq 3 \}}$, because we must choose an orientation of each undirected cycle in order to lift it to a permutation. Also, recall that $A_{ii}$ is even. So, if $\# \{i : a_i \geq 3 \} + \# \{i : a_i = 1 \}$ is at least $2$, then our monomial occurs with multiplicity divisible by $2^2 = 4$ and vanishes modulo $4$. So the only terms that survive are those with all but at most one of the $a_i$ equal to $2$.
Meanwhile, on the right hand side, we get those graphs which are a union of two perfect matchings. Such a graph must have all cycle lengths even. Assuming all the $a_i$ are even, the coefficient of such a term is $2^{\# \{i : a_i > 2 \}}$, such for each cycle of length $>2$, there are two choices for which edges will come from the first factor of $\text{haf}(A)$ and which from the second. So, again, any term with more than one $a_i$ greater than $2$ will drop out.
On both sides, we see that the only contributions come from terms with cycle decomposition of the form either $2+2+2+\cdots+2+2k$ for $k>1$, or $2+2+\cdots +2$. In the first case, the coefficient is $\pm 2$ on  both sides, so they match modulo $4$. In the second case, on the left hand side, we have $(-1)^m$, as that is the sign of a permutation with cycle type $2+2+\cdots +2$; this is also the coefficient on the right hand side. QED.

After writing up this answer, I did a little searching to find out where this question had come from. This result is Theorem 1 in "On Unimodular Graphs", Akbari and Kirkland (2007) and is Proposition 4.11 in "Modular forms and Hecke operators", Adrianov and Zhuravlev (1995).
