Let $N$ be the number of ordered 5-tuples $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ of positive integers satisfying

$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}+\frac{1}{a_{5}}=1$

Is $N$ even or odd?

Oh and [b]HINTS ONLY[/b], please do not give full solutions. Thanks.
1. **Understanding the problem**: We need to determine whether the number of ordered 5-tuples \((a_1, a_2, a_3, a_4, a_5)\) of positive integers satisfying the equation \(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \frac{1}{a_4} + \frac{1}{a_5} = 1\) is even or odd.

2. **Partitioning the problem**: Let \(N_{p_1, p_2, \ldots, p_k}\) (where \(p_1 \geq p_2 \geq \ldots \geq p_k\)) be the number of \((a_1, a_2, \ldots, a_5)\) that satisfy the equation such that the \(a_i\)'s can be partitioned into \(k\) groups. The \(a_i\)'s in the same group have identical values, but distinct from those in another group. For example, \((4, 4, 4, 8, 8)\) and \((8, 8, 4, 4, 4)\) count as one in \(N_{3,2}\).

3. **Summing over partitions**: We can write
   \[
   N = \sum_{\left(p_1, \ldots, p_k\right)} \binom{5}{p_1, \ldots, p_k} N_{p_1, \ldots, p_k}
   \]
   where \(\binom{5}{p_1, \ldots, p_k}\) is the multinomial coefficient.

4. **Focusing on parity**: To determine the parity of \(N\), we only need to compute \(N_5\) and \(N_{4,1}\). This is because the parity of the sum of all terms will be determined by the parity of these two terms.

5. **Computing \(N_5\)**: \(N_5\) corresponds to the case where all \(a_i\)'s are equal. In this case, we have:
   \[
   \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \frac{1}{a_4} + \frac{1}{a_5} = 1 \implies 5 \cdot \frac{1}{a_1} = 1 \implies a_1 = 5
   \]
   Thus, there is exactly one such 5-tuple: \((5, 5, 5, 5, 5)\). Therefore, \(N_5 = 1\).

6. **Computing \(N_{4,1}\)**: \(N_{4,1}\) corresponds to the case where four \(a_i\)'s are equal and the fifth one is different. Let \(a_1 = a_2 = a_3 = a_4 = x\) and \(a_5 = y\). Then we have:
   \[
   4 \cdot \frac{1}{x} + \frac{1}{y} = 1 \implies \frac{4}{x} + \frac{1}{y} = 1
   \]
   Solving for \(y\), we get:
   \[
   \frac{1}{y} = 1 - \frac{4}{x} \implies y = \frac{x}{x-4}
   \]
   For \(y\) to be a positive integer, \(x-4\) must divide \(x\), which implies \(x-4\) divides 4. The possible values for \(x\) are 5, 6, and 8. For each value of \(x\), we get a corresponding \(y\):
   - \(x = 5\), \(y = 5\) (not valid since \(y\) must be different from \(x\))
   - \(x = 6\), \(y = 6\) (not valid for the same reason)
   - \(x = 8\), \(y = 2\) (valid)

   Thus, there is exactly one valid 5-tuple: \((8, 8, 8, 8, 2)\). Therefore, \(N_{4,1} = 1\).

7. **Determining the parity**: Since both \(N_5\) and \(N_{4,1}\) are 1, which are odd, the sum \(N = N_5 + N_{4,1}\) is even.

The final answer is \(\boxed{\text{even}}\)