Can AES-CTR add values without decryption? I recently have studied AES cipher. On CTR mode, is it possible to add, subtract, or XOR plaintext values on encrypted data? 

For example: we have an encryption of $4$ and $6$ under the same AES key, respectively denoted by $AES(4)$ and $AES(6)$.

Without decryption, can we achieve the following?

$$AES(4)⊕AES(6)=AES(4+6)$$

Where "$\oplus$" is any operation we can do over AES.
No, with AES in CTR mode, it is not


  possible to add/subtract(or XOR?) plaintext values on encrypted data


nor define operator $\boxplus$ such that $\operatorname{AES-CTR}(4)\boxplus\operatorname{AES-CTR}(6)=\operatorname{AES-CTR}(4+6)$ that works for unknown key and plaintexts. Informally: if that worked, it would be a weakness of the encryption scheme, which would be leaking information about plaintexts from the ciphertexts.

Notice that the above notation is misleading, as the result of AES-CTR depends on 3 parameters: Key, IV and plaintext, with IV normally random (making the argument above non-rigorous). But still, even if we assume fixed IV, what's asked can't be done: it is asking an operator $\boxplus$ such that $\forall s,\forall x,\forall y, (s\oplus x)\boxplus(s\oplus y)=s\oplus(x+y)$ where $\oplus$ is XOR, and $s$ is whatever the IV encrypts to under the key considered; which is impossible.

More generally, no secure mode of operation of any secure block cipher allow homomorphic encryption.