Determining the half-reaction equations for extraordinary redox reaction?
You have rightfully assumed the reaction takes place in either basic or neutral medium, otherwise $\ce{Cr(OH)3}$ wouldn't be the product.
This automatically suggests $\ce{H+}$ are not used for balancing half reactions.
Your solution is fine for the provided reaction given that chromium(III) hydroxide stays precipitated.
Here I've basically formatted your solution in a form I'm more familiar with:

\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &amp;→ \overset{+3}{Cr}(OH)3 + 5 OH-} &amp;|\cdot 2 \tag{red1}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &amp;→ \overset{+6}{S}O4^2- + H2O + 2 e-} &amp;|\cdot 3 \tag{ox1}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 5 H2O &amp;→ 2 Cr(OH)3 + 3 SO4^2- + 4 OH-} \tag{redox1}
\end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 5 H2O → 2 Cr(OH)3 + 3 Na2SO4 + 4 NaOH} \tag{R1}$$

This implies that reaction begins in neutral medium; however there is remaining sodium hydroxide ($\ce{4 NaOH}$ on the right), so the pH will increase over time.
This will result in further dissolving of $\ce{Cr(OH)3}$ in $\ce{NaOH}$ yielding $\ce{Na3[Cr(OH)6]}$ complex, which is likely the final product.
Alternatively, addition of a base to this mix would also accelerate formation of chromium(III) hydroxocomplex.
With this in mind, I'd say the more appropriate way to write the reaction is

\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &amp;→ [\overset{+3}{Cr}(OH)6]^3- + 2 OH-} &amp;|\cdot 2 \tag{red2}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &amp;→ \overset{+6}{S}O4^2- + H2O + 2 e-} &amp;|\cdot 3 \tag{ox2}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 2 OH- + 5 H2O &amp;→ 2 [Cr(OH)6]^3- + 3 SO4^2-} \tag{redox2}
\end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 2 NaOH + 5 H2O → 2 Na3[Cr(OH)6] + 3Na2SO4} \tag{R2}$$