1. Let the set be $A=\{1,2,3,4\}.$ A relation on $A$ is a subset of $A\times A$.

2. For a relation to be symmetric, whenever $(a,b)$ is in the relation, $(b,a)$ must also be in the relation.

3. Count all symmetric relations on $A$.

• Diagonal pairs are: $(1,1),(2,2),(3,3),(4,4).$ Each of these can be chosen independently, since each is automatically symmetric. So number of choices for diagonal pairs is $2^4.$

• Off-diagonal pairs come in unordered symmetric pairs: $\{(1,2),(2,1)\},\{(1,3),(3,1)\},\{(1,4),(4,1)\},\{(2,3),(3,2)\},\{(2,4),(4,2)\},\{(3,4),(4,3)\}.$ There are $\binom{4}{2}=6$ such pairs. For each such pair, either both are included or both are excluded. So number of choices for off-diagonal part is $2^6.$

Hence total number of symmetric relations is $2^4\cdot 2^6=2^{10}=1024.$

4. Now subtract the symmetric relations that are reflexive.

A relation is reflexive if all diagonal pairs $(1,1),(2,2),(3,3),(4,4)$ are present.

So for reflexive symmetric relations:

• Diagonal entries are fixed: only $1$ choice.
• Off-diagonal symmetric pairs still have $2^6$ choices.

Thus number of reflexive symmetric relations is $2^6=64.$

5. Therefore, the number of symmetric relations that are not reflexive is $1024-64=960.$

So the required answer is $\boxed{960}.$