Here is an easy argument, answering affirmatively both questions. I have interpreted the slightly ambiguous wording as meaning that for each choice of $p,q$, with both positive (which means strictly positive), there exists a solution with all entries (strictly) positive.

Lemma. Given $q \geq 1$, there exists a tuple of positive integers, $(a; b_1, b_2, \dots, b_q)$ such that $a $ is odd and $a^2 = \sum b_i^2$.

Proof. By induction. If $q = 1$, set $a = b_1 =1$. Otherwise assume true for $q-1$; there exists a strictly positive tuple, $(r; b_1, b_2, \dots, b_{q-1}) $, such that $r^2 = \sum_{i=1}^{q-1} b_i^2$ with odd $r > 1 $. Then we can write $r^2 = ((r^2+1)/2)^2 - ((r^2 -1)/2)^2 =e^2 -f^2$, as a difference of squares, with $e$ odd, and $f$ even but nonzero. Set $a = e$ (so $a$ is odd) and $b_q = f$. Then $$\eqalign{ a^2 - \sum_1^q b_i^2 &= \left(r^2 - \sum_{i=1}^{q-1} b_i^2\right) + a^2 - r^2 - b_q^2\cr & = 0 + 0 = 0.\cr }$$

[Hypothesis about oddness is needed for the differences of squares argument.]

Corollary. If $p, q$ are positive integers, there exists a strictly positive integer solution $(x_1, \dots, x_p; y_1, \dots y_q) $ to the equation $$ \sum_{j=1}^p x_j^2 = \sum_{i=1}^q y_i^2. $$

Proof. Without loss of generality, $p \leq q$. If $p= q$, set all the variables equal to $1$. Otherwise, $p < q$; by setting $x_1 = x_2 = \dots= x_{p-1} = 1 = y_1 = y_2 = \dots = y_{p-1}$, we immediately reduce to the case that $p = 1$ and $q > 1$. But this is the conclusion of the lemma.

Presumably, by actually applying the induction argument with the smallest possible choices for $r$, one obtains small (or even the smallest) solutions (where we measure smallness by the sum)?

Here is an easy argument, answering affirmatively both questions. I have interpreted the slightly ambiguous wording as meaning that for each choice of $p,q$, with both positive (which means strictly positive), there exists a solution with all entries (strictly) positive.

Lemma. Given $q \geq 1$, there exists a tuple of positive integers, $(a; b_1, b_2, \dots, b_q)$ such that $a $ is odd and $a^2 = \sum b_i^2$.

Proof. By induction. If $q = 1$, set $a = b_1 =1$. Otherwise assume true for $q-1$; there exists a strictly positive tuple, $(r; b_1, b_2, \dots, b_{q-1}) $, such that $r^2 = \sum_{i=1}^{q-1} b_i^2$ with odd $r > 1 $. Then we can write $r^2 = ((r^2+1)/2)^2 - ((r^2 -1)/2)^2 =e^2 -f^2$, as a difference of square integers, with $e$ odd, and $f$ even but nonzero. Set $a = e$ (so $a$ is odd) and $b_q = f$. Then $$\eqalign{ a^2 - \sum_1^q b_i^2 &= \left(r^2 - \sum_{i=1}^{q-1} b_i^2\right) + a^2 - r^2 - b_q^2\cr & = 0 + 0 = 0.\cr }$$

[Hypothesis about oddness is needed for the differences of squares argument.]

Corollary. If $p, q$ are positive integers, there exists a strictly positive integer solution $(x_1, \dots, x_p; y_1, \dots y_q) $ to the equation $$ \sum_{j=1}^p x_j^2 = \sum_{i=1}^q y_i^2. $$

Proof. Without loss of generality, $p \leq q$. If $p= q$, set all the variables equal to $1$. Otherwise, $p < q$; by setting $x_1 = x_2 = \dots= x_{p-1} = 1 = y_1 = y_2 = \dots = y_{p-1}$, we immediately reduce to the case that $p = 1$ and $q > 1$. But this is the conclusion of the lemma.

Presumably, by actually applying the induction argument with the smallest possible choices for $r$, one obtains small (or even the smallest) solutions (where we measure smallness by the sum)?

The problem would be more difficult if all the entries of the solution were required to be distinct as well.

Here is an easy argument, answering affirmatively both questions. I have interpreted the slightly ambiguous wording as meaning that for each choice of $p,q$, with both positive (which means strictly positive), there exists a solution with all entries (strictly) positive.

Lemma. Given $q \geq 1$, there exists a tuple of positive integers, $(a; b_1, b_2, \dots, b_q)$ such that $a $ is odd and $a^2 = \sum b_i^2$.

Proof. By induction. If $q = 1$, set $a = b_1 =1$. Otherwise assume true for $q-1$; there exists a strictly positive tuple, $(r; b_1, b_2, \dots, b_{q-1}) $, such that $r^2 = \sum_{i=1}^{q-1} b_i^2$ with odd $r > 1 $. Then we can write $r^2 = ((r^2+1)/2)^2 - ((r^2 -1)/2)^2 =e^2 -f^2$, as a difference of squares, with $e$ odd, and $f \neq 0$. Set $a = e$ (so $a$ is odd) and $b_q = f$. Then $$\eqalign{ a^2 - \sum_1^q b_i^2 &= \left(r^2 - \sum_{i=1}^{q-1} b_i^2\right) + a^2 - r^2 - b_q^2\cr & = 0 + 0 = 0.\cr }$$

[Hypothesis about oddness is needed for the induction argument.]

Corollary. If $p, q$ are positive integers, there exists a strictly positive integer solution $(x_1, \dots, x_p; y_1, \dots y_q) $ to the equation $$ \sum_{j=1}^p x_j^2 = \sum_{i=1}^q y_i^2. $$

Proof. Without loss of generality, $p \leq q$. If $p= q$, set all the variables equal to $1$. Otherwise, $p < q$; by setting $x_1 = x_2 = \dots= x_{p-1} = 1 = y_1 = y_2 = \dots = y_{p-1}$, we immediately reduce to the case that $p = 1$ and $q > 1$. But this is the conclusion of the lemma.

Here is an easy argument, answering affirmatively both questions. I have interpreted the slightly ambiguous wording as meaning that for each choice of $p,q$, with both positive (which means strictly positive), there exists a solution with all entries (strictly) positive.

Lemma. Given $q \geq 1$, there exists a tuple of positive integers, $(a; b_1, b_2, \dots, b_q)$ such that $a $ is odd and $a^2 = \sum b_i^2$.

Proof. By induction. If $q = 1$, set $a = b_1 =1$. Otherwise assume true for $q-1$; there exists a strictly positive tuple, $(r; b_1, b_2, \dots, b_{q-1}) $, such that $r^2 = \sum_{i=1}^{q-1} b_i^2$ with odd $r > 1 $. Then we can write $r^2 = ((r^2+1)/2)^2 - ((r^2 -1)/2)^2 =e^2 -f^2$, as a difference of squares, with $e$ odd, and $f$ even but nonzero. Set $a = e$ (so $a$ is odd) and $b_q = f$. Then $$\eqalign{ a^2 - \sum_1^q b_i^2 &= \left(r^2 - \sum_{i=1}^{q-1} b_i^2\right) + a^2 - r^2 - b_q^2\cr & = 0 + 0 = 0.\cr }$$

[Hypothesis about oddness is needed for the differences of squares argument.]

Corollary. If $p, q$ are positive integers, there exists a strictly positive integer solution $(x_1, \dots, x_p; y_1, \dots y_q) $ to the equation $$ \sum_{j=1}^p x_j^2 = \sum_{i=1}^q y_i^2. $$

Proof. Without loss of generality, $p \leq q$. If $p= q$, set all the variables equal to $1$. Otherwise, $p < q$; by setting $x_1 = x_2 = \dots= x_{p-1} = 1 = y_1 = y_2 = \dots = y_{p-1}$, we immediately reduce to the case that $p = 1$ and $q > 1$. But this is the conclusion of the lemma.

Presumably, by actually applying the induction argument with the smallest possible choices for $r$, one obtains small (or even the smallest) solutions (where we measure smallness by the sum)?