$\def\dr{d_{\rightarrow}}\def\du{d_{\uparrow}}$ This is true. I don't know a reference, but here is a proof. I will show, more strongly, that, at every stage in the Hodge-de Rham spectral sequence, we have a perfect pairing $E^{pq}_r \times E^{(n-p)(n-q)}_r \to k$. In particular, $E^{pq}_{\infty}$ and $E^{(n-p)(n-q)}_{\infty}$ are dual and, since $H^k_{DR}(X)$ is filtered with associated graded $\bigoplus_{p+q=k} E^{pq}_{\infty}$, this shows that $H^k_{DR}$ and $H^{2n-k}_{DR}$ are dual.

We begin with a bunch of formal nonsense about spectral sequences.

Let $V_1$, $V_2$, $W_1$ and $W_2$ be $k$-vector spaces equipped with bilinear pairings $V_1 \times V_2 \to k$ and $W_1 \times W_2 \to k$.
Let $\phi_1 : V_1 \to W_1$ and $\phi_2 : W_2 \to V_2$ be linear maps. We say that $\phi_1$ and $\phi_2$ are adjoint if $\langle \phi_1(v_1), w_2 \rangle = \langle v_1, \phi_2(w_2) \rangle$ for all $v_1 \in V_1$ and $w_2 \in W_2$. (Note that, as yet, we don't require the pairings to be perfect or anything to be finite dimensional.)

Let $A^{pq}$ be an $n \times n$ double complex of $k$-vector spaces, with $\dr: A^{pq} \to A^{(p+1)q}$ and $\du: A^{pq} \to A^{p(q+1)}$ the rightward and upward maps. Set $d = \dr+\du$. (I'll be sloppy about signs throughout, but I think my convention is that squares anti-commute.)
Suppose that, for all $(p,q)$, we are given a bilinear pairing $A^{pq} \times A^{(n-p)(n-q)} \to k$; suppose that $\dr: A^{pq} \to A^{(p+1)q}$ and $\dr: A^{(n-p-1)(n-q)} \to A^{(n-p)(n-q)}$ are adjoint, as are $\du: A^{pq} \to A^{p(q+1)}$ and $\du: A^{(n-p)(n-q-1)} \to A^{(n-p)(n-q)}$. Let $E^{pq}_r$ be the corresponding spectral sequence, where we first take differentials in the $\du$ direction.

Our main results are

**Theorem 1:** The bilinear pairing between $A^{pq}$ and $A^{(n-p)(n-q)}$ descends to a bilinear pairing between the subquotients $E^{pq}_r$ and $E^{(n-p)(n-q)}_r$.
With respect to this pairing, the differentials $E^{pq}_r \to E^{(p+r)(q+1-r)}_r$ and $E^{(n-p-r)(n-q-1+r)}_r \to E^{(n-p)(n-q)}_r$ are adjoint.

**Theorem 2:** In the above setting, if the vector spaces on the $r$-th page are finite dimensional and the pairings between them are perfect, then so are the pairings on every succeeding page.

To prove this, we need to recall how $E^{pq}_r$ is defined. We follow Vakil, section 1.7.7, except for the unfortunate point that he does the map in the $p$ direction first and we do the map in the $q$ direction first. This is forced on us because the notation $H^q(X, \Omega^p)$ is standard. This means that many of our coordinates are reversed from Vakil.

Set $S^{pq} = \bigoplus_{k \geq 0} A^{(p+k)(q-k)}$. Let $S^{pq}_r$ be $S^{pq} \cap d^{-1}(S^{(p+r)(q+1-r)})$. (So $S^{pq}_0=S^{pq}$, and increasing $r$ makes the condition more restrictive.) Then
$$E^{pq}_r = \frac{S^{pq}_r}{S^{(p+1)(q-1)}_{r-1} + d S^{(p-r+1)(q+r-2)}_{r-1}}.$$
The term $S^{(p+1)(q-1)}_{r-1}$ is simply those elements in $S^{pq}_r$ which have no contribution from the $A^{pq}$ summand, so $S^{pq}_r/S^{(p+1)(q-1)}_{r-1}$ injects into $A^{pq}$ and $E^{pq}_r$ is a subquotient of $A^{pq}$.

**Lemma:** Let $p+p' = n-r$ and $q+q'=n+r-1$. Let $v \in S^{pq}_r$ and $w \in S^{p'q'}_r$. Then
$$\langle v^{pq}, (dw)^{(p'+r)(q'+1-r)} \rangle = \langle (d v)^{(p+r)(q+1-r)}, w^{p' q'} \rangle.$$
Here $u^{pq}$ denotes the projection of $u$ onto the $A^{pq}$ summand.

**Proof Sketch:** We have $(dw)^{(p'+r)(q'+1-r)} = \dr (w^{(p'-1+r)(q'+1-r)} ) + \du (w^{(p'+r)(q'-r)})$. By our adjointness hypotheses, $\langle v^{pq}, (dw)^{(p'+r)(q'+1-r)} \rangle = \langle \dr(v^{pq}), w^{(p'-1+r)(q'+1-r)} \rangle + \langle \du(v^{pq}), w^{(p'+r)(q'-r)} \rangle$. But $\du(v^{pq})=0$ since $v \in S^{pq}_r$, so we only need to think about the first term.
Now, since $v \in S^{pq}_r$, we have $\dr(v^{pq}) = - \du(v^{(p+1)(q-1)})$ and, using adjointness again, $\langle \du(v^{(p+1)(q-1)}), w^{(p'-1+r)(q'+1-r)} \rangle = \langle v^{(p+1)(q-1)}, \du (w^{(p'-1+r)(q'-r)}) \rangle$. Since $w \in S^{p'q'}_r$, we have $ \du (w^{(p'-1+r)(q'-r)}) = - \dr(w^{(p'-2+r, q'-r-1})$. Continuing in this manner, we eventually establish the result. $\square$

**Proof Sketch of Theorem 1:** We first check that the bilinear form descends to the quotient. In other words, if $v \in S^{(p+1)(q-1)}_{r-1} + d S^{(p-r+1)(q+r-2)}_{r-1}$ and $w \in S^{(n-p)(n-q)}_r$, then $\langle v^{pq} ,w^{(n-p)(n-q)} \rangle = 0$. If $v \in S^{(p+1)(q-1)}_{r-1}$ then $v^{pq}=0$, so this is immediate. Now, suppose that $v=du$ for $u \in S^{(p-r+1)(q+r-2)}_{r-1}$. Note that $S^{(p-r+1)(q+r-2)}_{r-1} \subseteq S^{(p-r)(q+r-1)}_r$. So the lemma shows that $\langle (du)^{pq}, w^{(n-p)(n-q)} \rangle = \langle u^{(p-r)(q+r-1)}, (dw)^{(n-p+r)(n-q-r+1)} \rangle$. But $u^{(p-r)(q+r-1)}=0$, since $u \in S^{(p-r+1)(q+r-2)}$. We have shown that the bilinear form descends.

We now recall the definition of the differential $d_r: E^{pq}_r \to E^{(p+r)(q+1-r)}_r$. Take $v \in E^{pq}_r$ and lift $v$ to $ \tilde{v} \in S^{pq}_r$. Then $d_r(v)$ is the class of $(d \tilde{v})^{(p+r)(q+1-r)}$ in the quotient $E^{(p+r)(q+1-r)}_{r}$. Take $v \in E^{pq}_r$ and $w \in E^{(n-p-r)(n-q-1+r)}_r$ and lift them to $\tilde{v}$ and $\tilde{w}$. We want to show that $\langle (d \tilde{v})^{(p+r)(q+1-r)}, \tilde{w}^{(n-p-r)(n-q-1-r)} \rangle = \langle \tilde{v}^{pq}, (d \tilde{w})^{(n-p)(n-q)} \rangle$. Again, this is the Lemma. $\square$

**Proof Sketch of Theorem 2:** Suppose that all the vector spaces on the $r$-th page are finite dimensional and the pairings between them are perfect. Write $V^{\vee}$ for the dual to a finite dimensional vector space $V$. We have complexes $\cdots \to E^{(p-r)(q+r-1)}_r \to E^{pq}_r \to E^{(p+r)(q-r+1)}_r \to \cdots$, and $E^{pq}_{r+1}$ is the cohomology of this complex. If two complexes of finite dimensional vector spaces are dual, then their cohomologies are also dual. "$\square$"

Now, we must explain why all of this applies to our setting. Let $X$ be smooth projective irreducible of dimension $n$ over a field $k$. We can extend the base field, and hence can assume the base field is infinite. Thus, by Bertini, we can find $n+1$ mutually transverse very ample divisors $D_0$, $D_1$, \ldots, $D_n$. Let $U(i_0, i_1, \ldots, i_q)$ be the affine open $X \setminus (D_{i_0} \cup \cdots \cup D_{i_q})$. Set $A^{pq} = \bigoplus_{0 \leq i_0 < i_1 < \cdots < i_q \leq n} \Omega^p(U(i_0, \ldots, i_q))$.
We organize these into a double complex in the usual way, so $\dr$ is the de Rham differential and $\du$ is the Cech differential.
$H^{\ast}_{DR}$ is the total cohomology of this complex.

We note that $A^{nn}$ is Cech co-chains for $H^n(X, \Omega^n)$ and, since we only have $n+1$ open sets, these co-chains are co-cycles. So we have a natural projection $A^{nn} \to H^n(X, \Omega^n) \cong k$, by Serre duality. We denote this map $A^{nn} \to k$ by $\int$.

We have a product $A^{pq} \times A^{p' q'} \to A^{(p+p')(q+q')}$ by $(\alpha \beta)(i_0, i_1, \ldots, i_{q+q'}) = \alpha(i_0, i_1, \ldots, i_q) \wedge \beta(i_q, \ldots, i_{q+q'})$. (Being sloppy about signs as usual, and omitting notation for restriction.
It is standard (and also easy, if you neglect signs) that $\dr$ and $\du$ are both derivations for this multiplication.
We define a pairing $A^{pq} \times A^{(n-p)(n-q)} \to k$ by $\langle \alpha , \beta \rangle = \int \alpha \beta$.

We wish to check that $\dr$ and $\du$ are self-adjoint. We have $\langle \du(\alpha), \beta \rangle = \int \du(\alpha) \beta$ and $\langle \alpha, \du(\beta) \rangle = \int \alpha \du(\beta)$. So our goal is to show that $\int \alpha \du(\beta) \pm \du(\alpha) \beta = 0$ or, in other words, that $\int \du(\alpha \beta)=0$. This just says that a Cech coboundary is zero in $H^n(X, \Omega^n)$, so it is true.
When we do the same computation for $\dr$, we wind up needing to show that $\int \dr(\alpha \beta)=0$. The explicit description of $\int$ in terms of residues makes it clear that $\int d(\eta)=0$ for any $\eta \in A^{(n-1)n}$.

So the hypotheses of our general set up apply. Also, $E^{pq}_1 \cong H^q(X, \Omega^p)$, so the hypotheses of Theorem 2 apply for $r=1$ by Serre duality and the finiteness of cohomology of coherent sheaves. Theorem 2 then proves the result.