1. We check the thermal decomposition of each compound and see whether $N_2$ gas is formed.

2. Option A: $(NH_4)_2SO_4$

On heating, ammonium sulfate does not give nitrogen gas. It mainly decomposes to products such as ammonia and acidic sulfate-related species, not $N_2$.

So, this option does not produce $N_2$.

3. Option B: $Ba(N_3)_2$

$N_3^-$ is the azide ion. Metal azides decompose on heating to give the metal and nitrogen gas:

$Ba(N_3)_2 \rightarrow Ba + 3N_2$

So, this option does produce $N_2$.

4. Option C: $(NH_4)_2Cr_2O_7$

Ammonium dichromate decomposes as:

$(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + N_2 + 4H_2O$

So, this option does produce $N_2$.

5. Option D: $NH_4NO_2$

Ammonium nitrite decomposes as:

$NH_4NO_2 \rightarrow N_2 + 2H_2O$

So, this option does produce $N_2$.

6. Therefore, the only compound that does not produce nitrogen gas on thermal decomposition is:

$\boxed{(NH_4)_2SO_4}$

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, the derived answer agrees with the stored correct answer.