Let me start by saying that your question is definitely not a research-level question --- usually similar questions are asked by first/second year undergraduate students in computer science.

However, taking into account the popularity of your question, let me give two remarks.

A) "All partially recursive problems are recursive". Here is a "proof". Let me fix a partially recursive problem that is hard in the class of partially recursive functions --- for example, validity in First-Order Logic (the argument is exactly the same for any partially recursive or corecursive problem). There is an obvious partial algorithm $A$ that solves the problem --- given a formula $\phi$ it explores the derivation tree for $\phi$ in breadth-first order in a (sound and complete) proof system for FOL. Let us assume that $\phi$ is valid. By completeness of the system, for every valid formula $\phi$, there exists the shortest proof $s(\phi)$ of $\phi$ in our system. Let us denote by $l(\phi)$ the length of the proof $s(\phi)$.

The trick is to modify algorithm $A$ by introducing a counter that counts the length of our search tree --- whenever the value of the counter excesses $l(\phi)$, we break the computation process and return answer "no".

Of course, function $l$ may be uncomputable, but that is not a problem --- it suffices to bound $l$ from the above by any computable function. Functions like:
$$n^n$$
$$\left.n^{n^{\cdots^n}}\right\}n$$
$$\mathit{Ackermann}(n, n)$$
grow incredibly fast, and, together with their compositions, are clearly computable. We can just carefully choose one of such functions relatively to our proof system.

(The gap in the above "proof" is in the last statement --- the length of the shortest proof of a formula, grows asymptotically faster than any computable function.)

The lesson that should be taken from the above considerations, is that the true difficulty of solving uncomputable problems is not in the fact that we cannot design an algorithm that always terminates, but in the fact that the "computational complexity" of these problems is just enormously high --- in some sense the cost of solving these problems is so big, that we cannot effectively separate it from infinity (and so, sometimes our algorithms fail to halt at all).

When people come first to the concept of computability, they usually regret the fact that some problems are uncomputable, and try to "repair" this situation --- typically by "completing" the concept of computability, or in other terms, by making the negation commute with the computation. The crucial thing to understand here is that there is nothing to regret, and that the idea of "repairing" the situation is completely crazy --- uncomputable problems are just complex in their nature.

B) One may formalize your question in the following way. Let us say that a problem $f$ is in $P/Q$ if there exists a sequence $A_1, A_2, \dotsc$ of algorithms with the following property:

- $A_k$ collectively solves $f$ in time $P$, that is --- there exists a function $p \in P$, such that for every natural $k$ and for every $x$ such that $|x| \leq k$ algorithm $A_k$ halts on $x$ in at most $p(k)$ steps and $A_k(x) = f(x)$
- sizes of algorithms $A_k$ are bounded by $Q$, that is --- there exists a function $q \in Q$ such that the size of $A_k$ is at most $q(k)$.

(In fact, the above is a hipster-way to define the non-uniform complexity hierarchy $P/Q$.)

Now, your question may be rephrased --- is the halting problem in $P/Q$, for any reasonable classes $P$ and $Q$? The answer to this question is no (assuming the natural coding of algorithms; otherwise there are silly examples --- i.e. the halting problem under unary coding is obviously in $n/\mathit{Poly}$, since every problem is in $n/\mathit{Exp}$).